The key graphical properties of quadratic functions and parabolas were covered in List 9.2.13. If we don't have the graph of a quadratic function though, how do we know where these key features occur? In this section we'll explore how to algebraically determine where these features occur. We'll then use these processes to graph quadratic functions.
Let's start by looking at a somewhat complicated quadratic function that models the path of a baseball. The height of the baseball, \(H\) (measured in feet), after \(t\) seconds is given by \(H=-16t^2+75t+4.25\text{.}\) What information about this function is important? In general, we want to know the initial height, the maximum height, and the amount of time it takes for the ball to hit the ground. These key features will correspond to the vertical intercept, the vertex, and the horizontal intercept(s).
The graph of this function is shown in Figure 9.3.2.
Notice that with this function, we cannot easily read where either the vertex or the intercepts occur. To do so we'll need algebraic methods for determining where these key features occur.
As we've seen, parabolas can open either upward or downward. A simple example of a parabola that open upward is \(y=x^2\text{,}\) and a simple example of a parabola that opens downward is \(y=-x^2\text{:}\)
The sign of the leading coefficient on a quadratic function determines whether the graph of this function opens upward or downward. If the leading coefficient is a positive number, the parabola opens upward. If the leading coefficient is a negative number, the parabola opens downward.
To see why the leading coefficient of a parabola makes it open upward or downward, let's look numerically at values of the quadratic functions that differ only by the sign of their leading coefficients, \(y=x^2+2x+3\) and \(y=-x^2+2x+3\text{:}\)
| \(x\) | \(y=x^2+2x+3\) |
| \(-100\) | \(\highlight{{}(-100)^2{}}+2(-100)+3=\highlight{{}9803{}}\) |
| \(-10\) | \(\highlight{{}(-10)^2{}}+2(-10)+3=\highlight{{}83{}}\) |
| \(-3\) | \((-3)^2+2(-3)+3=6\) |
| \(-2\) | \((-2)^2+2(-2)+3=3\) |
| \(-1\) | \((-1)^2+2(-1)+3=2\) |
| \(0\) | \((0)^2+2(0)+3=3\) |
| \(1\) | \((1)^2+2(1)+3=6\) |
| \(2\) | \((2)^2+2(2)+3=11\) |
| \(3\) | \((3)^2+2(3)+3=18\) |
| \(10\) | \(\highlight{{}(10)^2{}}+2(10)+3=\highlight{{}123{}}\) |
| \(100\) | \(\highlight{{}(100)^2{}}+2(10)+3=\highlight{{}10203{}}\) |
| \(x\) | \(y=-x^2+2x+3\) |
| \(-100\) | \(\highlight{{}-(-100)^2{}}+2(-100)+3=\highlight{{}-10197{}}\) |
| \(-10\) | \(\highlight{{}-(-10)^2{}}+2(-10)+3=\highlight{{}-117{}}\) |
| \(-3\) | \(-(-3)^2+2(-3)+3=-12\) |
| \(-2\) | \(-(-2)^2+2(-2)+3=-5\) |
| \(-1\) | \(-(-1)^2+2(-1)+3=0\) |
| \(0\) | \(-(0)^2+2(0)+3=3\) |
| \(1\) | \(-(1)^2+2(1)+3=4\) |
| \(2\) | \(-(2)^2+2(2)+3=3\) |
| \(3\) | \(-(3)^2+2(3)+3=0\) |
| \(10\) | \(\highlight{{}-(10)^2{}}+2(10)+3=\highlight{{}-77{}}\) |
| \(100\) | \(\highlight{{}-(100)^2{}}+2(10)+3=\highlight{{}-9797{}}\) |
As we can see in Table 9.3.6 and Table 9.3.7, for large positive values of \(x\) and for large negative values of \(x\) the \(x^2\) term dominates. This is why its coefficient will determine whether the parabola opens upward or downward.
Determine if the graph of each quadratic function opens upward or downward.
In List 9.2.13, we identified that horizontal intercepts occur where the graph of a function intersects the horizontal axis and that the vertical intercept occurs where the graph of a function intersects the vertical axis. If we're using \(x\) and \(y\) as our variables, this means that the \(x\)-intercept(s) occur where \(y=0\text{,}\) and the \(y\)-intercept occurs where \(x=0\text{.}\) These might seem the reverse of what you'd expect, but it's important to remember that when finding the \(\highlight{{}x{}}\)-intercept, we want to determine the value(s) of \(\highlight{{}x{}}\) when \(y\) is set to zero. Similarly, when determining the \(\highlight{{}y{}}\)-intercept, we want to determine the value of \(\highlight{{}y{}}\) when \(x\) is zero. We'll now look at determining these algebraically.
Algebraically determine the \(x\)-intercept(s) and \(y\)-intercept for the quadratic function \(y=x^2-4x-12\text{.}\)
To determine the \(x\)-intercept(s), we will set \(y=0\) and solve for \(x\text{:}\)
\begin{align*} y\amp=x^2-4x-12\\ \substitute{0}\amp=x^2-4x-12\\ 0\amp=(x-6)(x+2)\\ x\amp-6=0,\ x+2=0\\ x\amp=6,\ x=-2 \end{align*}So the \(x\)-intercepts occur where \(x=6\) and where \(x=-2\text{.}\) On a graph, these associate with the points \((6,0)\) and \((-2,0)\text{.}\)
To determine the \(y\)-intercept, we will set \(x=0\) and find \(y\text{:}\)
\begin{align*} y\amp=x^2-4x-12\\ y\amp=\substitute{0}^2-4(\substitute{0})-12\\ y\amp=-12 \end{align*}So the \(y\)-intercept occurs where \(y=-12\text{.}\) On a graph, this associates with the point \((0,-12)\text{.}\)
The vertex of a quadratic function is either the minimum point or the maximum point on its graph, depending on whether this function opens upward or downward. When there are two horizontal intercepts for a quadratic function, we can see that the vertex occurs halfway between those two points. Similarly, when there is exactly one horizontal intercept, we know that this point is where the vertex occurs.
This visual method is quick, but it can only be used in limited cases. We cannot use this method if there are no horizontal intercepts, and it's also challenging to use if there are two horizontal intercepts with \(x\)-values that are irrational numbers.
The vertex of a quadratic function \(f(x)=ax^2+bx+c\) occurs where \(x=-\frac{b}{2a}\text{,}\) or at the point \(\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)\text{.}\)
This formula may at first seem a bit random, but it is actually related to the quadratic formula. Still thinking about the vertex occuring in the middle of two horizontal intercepts, we can see that this value is in the middle of the two values obtained from the quadratic formula:
\begin{equation*} x=\frac{{}\highlight{-b}{}\pm \lowlight{\sqrt{b^2-4ac}}}{{}\highlight{2a}{}} \end{equation*}Algebraically determine the vertex of the quadratic function \(y=x^2-4x-12\text{.}\)
To algebraically determine this vertex, we could rely on the horizontal intercepts determined in Example 9.3.9. To do so, we'll note that the vertex and axis of symmetry will occur halfway between \(x=6\) and \(x=-2\text{,}\) which is at \(x=2\text{.}\) Visually, this looks like:
We could also determine the \(x\)-value where the vertex occurs using the formula \(x=-\frac{b}{2a}\text{.}\) Noting that for \(y=x^2-4x-12\) we have \(a=1\) and \(b=-4\text{,}\) so:
\begin{align*} x\amp=-\frac{b}{2a}\\ x\amp=-\frac{\substitute{-4}}{2(\substitute{1})}\\ x\amp=2 \end{align*}Now that we've determined that the \(x\)-value where the vertex occurs is \(2\text{,}\) we will replace \(x\) with \(2\) in the original equation to determine \(y\text{:}\)
\begin{align*} y\amp=x^2-4x-12\\ y\amp=(\substitute{2})^2-4(\substitute{2})-12\\ y\amp=4-8-12\\ y\amp=-16 \end{align*}The vertex is the point \((2,-16)\) and the axis of symmtery is the line \(x=2\text{.}\)
To graph a quadratic function using its key features, we'll want to algebraically determine the following: whether the function opens upward or down, the horizontal intercepts, the vertical intercept, and the vertex. We'll then use this to connect the key points with a smooth curve.
Graph the function \(y=2x^2+10x+8\) by algebraically determining its key features.
To start, we'll note that this function will open upward, as the leading coefficient is the positive number \(2\text{.}\)
Next, we'll find the horizontal intercepts by setting \(y=0\) and solving for \(x\text{:}\)
\begin{align*} 2x^2+10x+8\amp=0\\ 2(x^2+5x+4)\amp=0\\ 2(x+4)(x+1)\amp=0\\ x+4\amp=0,\ x+1=0\\ x\amp=-4,\ x=-1 \end{align*}The \(x\)-intercepts are then \((-4,0)\) and \((-1,0)\text{.}\)
To find the \(y\)-intercept, we'll replace \(x\) with \(0\text{:}\)
\begin{align*} y\amp=2(\substitute{0})^2+10(\substitute{0})+8\\ y\amp=8 \end{align*}The \(y\)-intercept is then \((0,8)\text{.}\)
Lastly, we'll determine the vertex. Noting that \(a=2\) and \(b=10\text{,}\) we have:
\begin{align*} x\amp=-\frac{10}{2(2)}\\ x\amp=-2.5 \end{align*}Using this \(x\)-value to find the \(y\)-coordinate, we have:
\begin{align*} y\amp=2(\substitute{-2.5})^2+10(\substitute{-2.5})+8\\ y\amp=12.5-25+8\\ y\amp=-4.5 \end{align*}The vertex is then the point \((-2.5,-4.5)\text{,}\) and the axis of symmetry is the line \(x=-2.5\text{.}\)
We're now ready to graph this function. We'll start by drawing and scaling axes so that all of our key features will be displayed as shown in Figure 9.3.17. Next, we'll plot these key points as shown in Figure 9.3.18. Finally, we'll note that this parabola opens upward and connect these points with a smooth curve, as shown in Figure 9.3.19.
Graph the function \(y=-x^2+4x-5\) by algebraically determining its key features.
To start, we'll note that this function will open downward, as the leading coefficient is the negative number \(-1\text{.}\)
Next, we'll find the horizontal intercepts by setting \(y=0\) and solving for \(x\text{.}\) As we cannot use factoring to solve this equation, we'll use the quadratic formula:
\begin{align*} -x^2+4x-5\amp=0\\ x\amp=\frac{-4\pm \sqrt{(4)^2-4(-1)(-5)}}{2(-1)}\\ x\amp=\frac{-4\pm \sqrt{16-20}}{-2}\\ x\amp=\frac{-4\pm \sqrt{-8}}{-2} \end{align*}This shows that there are no real solutions to the equation, thus there are no horizontal intercepts on the graph of \(y=-x^2+4x-5\text{.}\)
To find the \(y\)-intercept, we'll replace \(x\) with \(0\text{:}\)
\begin{align*} y\amp=-(\substitute{0})^2+4(\substitute{0})-5\\ y\amp=-5 \end{align*}The \(y\)-intercept is then \((0,-5)\text{.}\)
To determine the vertex, we'll use \(a=-1\) and \(b=4\text{:}\)
\begin{align*} x\amp=-\frac{4}{2(-1)}\\ x\amp=2 \end{align*}Using this \(x\)-value to find the \(y\)-coordinate, we have:
\begin{align*} y\amp=-(\substitute{2})^2+4(\substitute{2})-5\\ y\amp=-4+8-5\\ y\amp=-1 \end{align*}The vertex is then the point \((2,-1)\text{,}\) and the axis of symmetry is the line \(x=2\text{.}\)
Plotting this information in an appropriate grid, we have:
As we can see in <<Unresolved xref, reference "figure-quadratic-function-no-horizontal-intecepts-key-features-initial-step-for-graphing-additional-point"; check spelling or use "provisional" attribute>>, this might be enough information for us to sketch the graph. If not, we'll first create a table to find a few more values around the vertex, add these, and then draw a smooth curve. This is shown in Table 9.3.24 and Figure 9.3.25.
| \(x\) | \(y=-x^2+4x-5\) | Point |
| \(-1\) | \(-(-1)^2+4(-1)-5=-10\) | \((-1,-10)\) |
| \(1\) | \(-(1)^2+4(1)-5=-2\) | \((1,-2)\) |
| \(3\) | \(-(3)^2+4(3)-5=-2\) | \((3,-2)\) |
| \(5\) | \(-(5)^2+4(5)-5=-10\) | \((5,-10)\) |
Returning to Example 9.3.9, we can now algebraically determine the horizontal and vertical intercepts and the vertex of the quadratic function \(H=-16t^2+75t+4.25\text{.}\) After doing so, we will interpret what these values mean in the context of this model.
To determine the horizontal intercepts, we'll solve \(H=0\text{.}\) Since factoring is not a possibility to solve this equation, we'll rely on the quadratic formula:
\begin{align*} H\amp=0\\ -16t^2+75t+4.25\amp=0\\ t\amp=\frac{-75\pm \sqrt{75^2-4(-16)(4.25)}}{2(-16)}\\ t\amp=\frac{-75\pm \sqrt{5897}}{-32}\\ \end{align*}Rounding these two values with a calculator, we obtain:
\begin{align*} t\amp\approx -0.056,\ t\approx 4.743 \end{align*}The horizontal intercepts therefore occur at approximately \((-0.056,0)\) and \((4.743,0)\text{.}\)
To determine the vertical intercept, we'll find \(H\) when \(t=0\text{:}\)
\begin{align*} H\amp=-16t^2+75t+4.25\\ H\amp=-16(\substitute{0})^2+75(\substitute{0})+4.25\\ H\amp=4.25 \end{align*}The vertical intercept therefore occurs at \((0,4.25)\text{.}\)
The vertex occurs where \(t=-\frac{b}{2a}\text{,}\) and for this function \(a=-16\) and \(b=75\text{.}\) So we have:
\begin{align*} t\amp=-\frac{75}{2(-16)}\\ t\amp=2.34375 \end{align*}We can now find the output for this input:
\begin{align*} H\amp=-16(\substitute{2.34375})^2+75(\substitute{2.34375})+4.25\\ \amp=92.140625 \end{align*}Thus the vertex is \((2.34375,92.140625)\text{.}\)
In context, the second horizontal intercept tells us that the height of the baseball was 0 ft after about 5.2692 s. If we assume this model begins when \(t=0\text{,}\) then the first horizontal intercept found is irrelevant.
The vertical intercept tells us that the initial height of the baseball was 4.25 ft. The vertex tells us that the maximum height was 92.140625 ft and occurred after 2.34375 s.
The profit that a manufacturing company makes for selling \(n\) units of refrigerators is given by \(P=-0.01n^2+520n-54000\text{.}\)
Determine the maximum profit and the number of refrigerators sold that yield this profit.
How many refrigerators need to be sold in order for the company to “break even”? (In other words, for their profit to be \(\$0\text{.}\))
The vertex of this function will tell us both the maximum profit and how many units are sold to yield this profit. Using \(a=-0.01\) and \(b=520\text{,}\) we have:
\begin{align*} n\amp=-\frac{b}{2a}\\ \amp=-\frac{520}{2(-0.01)}\\ \amp=26000 \end{align*}Now we will find the value of \(P\) when \(n=26000\text{:}\)
\begin{align*} P\amp=-0.01(\substitute{26000})^2+520(\substitute{26000})-54000\\ \amp=6706000 \end{align*}The maximum profit is \(\$6{,}706{,}000\text{,}\) which occurs if \(26{,}000\) units are sold.
To find how many refrigerators need to be sold in order for the company to “break even”, we will set \(P=0\) and solve for \(n\) using the quadratic formula:
\begin{align*} 0\amp=-0.01n^2+520n-54000\\ n\amp=\frac{-520\pm \sqrt{520^2-4(-0.01)(-54000)}}{2(-0.01)}\\ n\amp\frac{-520\pm\sqrt{268240}}{-0.02}\\ n\amp\approx 104, n\approx 51896 \end{align*}The company will break even if they sell about \(104\) refrigerators or \(51{,}896\) refrigerators.
Algebraically determine any horizontal and vertical intercepts and the vertex of the quadratic function \(y=-x^2+5x-7\text{.}\)
To determine the horizontal intercepts, we'll set \(y=0\) and solve for \(x\text{:}\)
\begin{align*} 0\amp=-x^2+5x-7\\ \end{align*}Noting that this equation cannot be solved using factoring, we'll use the quadratic formula:
\begin{align*} x\amp=\frac{-5\pm \sqrt{5^2-4(-1)(-7)}}{2(-1)}\\ x\amp=\frac{-5\pm \sqrt{25-28}}{-2}\\ x\amp=\frac{-5\pm \sqrt{-3}}{-2} \end{align*}Since these numbers are not real, no solution exists to the equation and so there are no horizontal intercepts.
To determine the vertical intercept, we'll replace \(x\) with \(0\text{:}\)
\begin{align*} y\amp=-(0)^2+5(0)-7\\ y\amp=7 \end{align*}Thus the \(y\)-intercept occurs at the point \((0,-7)\text{.}\)
The vertex will occur at \(x=-\frac{b}{2a}\text{:}\)
\begin{align*} x\amp=-\frac{5}{2(-1)}\\ \amp=\frac{5}{2} \end{align*}To find the \(y\)-coordinate, we'll replace \(x\) with \(\frac{5}{2}\text{:}\)
\begin{align*} y\amp=-\left(\substitute{\frac{5}{2}}\right)^2+5\left(\substitute{\frac{5}{2}}\right)+7\\ \amp=-\frac{3}{4} \end{align*}The vertex occurs at the point \(\left(\frac{5}{2},-\frac{3}{4}\right)\text{,}\) or at \((2.5,-0.75)\text{.}\)
Algebraically determine any horizontal and vertical intercepts and the vertex of the quadratic function \(y=-2(x+5)^2+12\text{.}\)
To determine the horizontal intercepts, we'll set \(y=0\) and solve for \(x\text{:}\)
\begin{align*} 0\amp=-2(x+5)^2+12\\ \end{align*}We cannot immediately factor this equation, but we can use the square root property to solve it. To do so, we'll first need to subtract \(12\) from each side of the equation and then divide each side by \(-2\text{:}\)
\begin{align*} -12\amp=-2(x+5)^2\\ 6\amp=(x+5)^2\\ \end{align*}Now applying the square root property,
\begin{align*} \pm\sqrt{6}\amp=x+5\\ x\amp=-\sqrt{6}-5,\ x=\sqrt{6}-5\\ x\amp\approx -7.449,\ x\approx -2.551 \end{align*}The \(x\)-intercepts occur at \((-\sqrt{6}-5,0)\) and \((\sqrt{6}-5,0)\text{,}\) or at approximately \((-7.449,0)\) and \((-2.551,0)\text{.}\)
To determine the \(y\)-intercept, we'll replace \(x\) with \(0\text{:}\)
\begin{align*} y\amp=-2(\substitute{0}+5)^2+12\\ \amp=-2(25)+12\\ \amp=-38 \end{align*}The \(y\)-intercept occurs at \((0,-38)\text{.}\)
To determine the vertex using the formula \(x=-\frac{b}{2a}\text{,}\) we'll first need to expand this equation so that \(a\) and \(b\) can be identified:
\begin{align*} y\amp=-2(x+5)^2+12\\ y\amp=-2(x^2+10x+25)+12\\ y\amp=-2x^2-20x-50+12\\ y\amp=-2x^2-20x-38 \end{align*}Now we can use \(a=-2\) and \(b=-20\) to determine the \(x\)-value of the vertex:
\begin{align*} x\amp=-\frac{-20}{2(-2)}\\ \amp=-5 \end{align*}Lastly, we'll use this value to find the \(y\)-value. We can use either the original equation or the expanded one; we'll use the original one here:
\begin{align*} y\amp=-2(\substitute{-5}+5)^2+12\\ \amp=-2(0)+12\\ \amp=12 \end{align*}Therefore the vertex is the point \((-5,12)\text{.}\)
Algebraically determine any horizontal and vertical intercepts and the vertex of the quadratic function \(y=5x^2-2x-\frac{1}{3}\text{.}\)
To determine the horizontal intercepts, we'll solve \(0=5x^2-2x-\frac{1}{3}\) using the quadratic formula. One option is to directly use \(a=5\text{,}\) \(b=-2\text{,}\) and \(c=-\frac{1}{3}\text{.}\) Another option is to first multiply each side of that equation by \(3\text{,}\) which will yield the equivalent equation \(0=15x^2-6x-1\text{.}\) From there, we can use the quadratic formula with \(a=10\text{,}\) \(b=-6\) and \(c=-1\text{:}\)
\begin{align*} 0=\amp=5x^2-2x-\frac{1}{3}\\ \multiplyleft{3}(0)\amp=\multiplyleft{3}\left(5x^2-2x-\frac{1}{3}\right)\\ 0\amp=15x^2-6x-1\\ x\amp=\frac{-(-6)\pm \sqrt{(-6)^2-4(15)(-1)}}{2(15)}\\ x\amp=\frac{6\pm \sqrt{96}}{30}\\ \end{align*}Simplifying, we get:
\begin{align*} x\amp=\frac{6\pm 4\sqrt{6}}{30}\\ x\amp=\frac{3\pm 2\sqrt{6}}{15}\\ \end{align*}Rouding, we have:
\begin{align*} x\amp\approx -0.1266,\ x\approx 0.5266 \end{align*}The horizontal intercepts occur at the points \(\left(\frac{3+ 2\sqrt{6}}{15},0\right)\) or \(\left(\frac{3-2\sqrt{6}}{15},0\right)\) , or at the approximate points \((-0.1266,0)\) and \((0.5266,0)\text{.}\)
To determine the vertical intercept, we'll set \(x=0\text{:}\)
\begin{align*} y\amp=5(\substitute{0})^2-2(\substitute{0})-\frac{1}{3}\\ \amp=-\frac{1}{3} \end{align*}Thus the \(y\)-intercept is \(\left(0,-\frac{1}{3}\right)\text{.}\)
Using \(a=5\) and \(b=-2\text{,}\) we see that the vertex occurs at:
\begin{align*} x\amp=-\frac{\substitute{-2}}{2(\substitute{5})}\\ \amp=\frac{1}{5} \end{align*}Now using this to determine the \(y\)-value, we have:
\begin{align*} y\amp=5\left(\substitute{\frac{1}{5}}\right)^2-2\left(\substitute{\frac{1}{5}}\right)-\frac{1}{3} \end{align*}Exercises on Algebraically Determining the Vertex and Axis of Symmetry of Quadratic Functions
Exercises on Algebraically Determining the Intercepts of Quadratic Functions
Sketching Graphs of Quadratic Functions
Graph \(y=x^2-7x+12\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((3,0)\) and \((4,0)\text{.}\)
The \(y\)-intercept is \((0,12)\text{.}\)
The vertex is \((3.5,-0.25)\) and the axis of symmetry is \(x=3.5\text{.}\)
Symmetry can be used to determine the point \((7,12)\text{.}\)
Graph \(y=x^2+5x-14\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-7,0)\) and \((2,0)\text{.}\)
The \(y\)-intercept is \((0,-14)\text{.}\)
The vertex is \((-2.5,-20.25)\) and the axis of symmetry is \(x=-2.5\text{.}\)
Symmetry can be used to determine the point \((-5,-14)\text{.}\)
Graph \(y=-x^2-x+20\) by algebraically determining its key features.
The parabola opens downward.
The \(x\)-intercepts are \((-5,0)\) and \((4,0)\text{.}\)
The \(y\)-intercept is \((0,20)\text{.}\)
The vertex is \((-0.5,20.25)\) and the axis of symmetry is \(x=-0.5\text{.}\)
Symmetry can be used to determine the point \((-1,20)\text{.}\)
Graph \(y=-x^2+4x+21\) by algebraically determining its key features.
The parabola opens downward.
The \(x\)-intercepts are \((-3,0)\) and \((7,0)\text{.}\)
The \(y\)-intercept is \((0,21)\text{.}\)
The vertex is \((2,25)\) and the axis of symmetry is \(x=2\text{.}\)
Symmetry can be used to determine the point \((4,21)\text{.}\)
Graph \(y=x^2-8x+16\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercept is \((4,0)\text{.}\)
The \(y\)-intercept is \((0,16)\text{.}\)
The vertex is \((4,0)\) and the axis of symmetry is \(x=4\text{.}\)
Symmetry can be used to determine the point \((8,16)\text{.}\)
Graph \(y=x^2+6x+9\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercept is \((-3,0)\text{.}\)
The \(y\)-intercept is \((0,9)\text{.}\)
The vertex is \((-3,0)\) and the axis of symmetry is \(x=-3\text{.}\)
Symmetry can be used to determine the point \((-6,9)\text{.}\)
Graph \(y=x^2-4\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-2,0)\) and \((2,0)\text{.}\)
The \(y\)-intercept is \((0,-4)\text{.}\)
The vertex is \((0,-4)\) and the axis of symmetry is \(x=0\text{.}\)
Graph \(y=x^2-9\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-3,0)\) and \((3,0)\text{.}\)
The \(y\)-intercept is \((0,-9)\text{.}\)
The vertex is \((0,-9)\) and the axis of symmetry is \(x=0\text{.}\)
Graph \(y=x^2+6x\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-6,0)\) and \((0,0)\text{.}\)
The \(y\)-intercept is \((0,0)\text{.}\)
The vertex is \((-3,-9)\) and the axis of symmetry is \(x=-3\text{.}\)
Graph \(y=x^2-8x\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((0,0)\) and \((8,0)\text{.}\)
The \(y\)-intercept is \((0,0)\text{.}\)
The vertex is \((4,-16)\) and the axis of symmetry is \(x=4\text{.}\)
Graph \(y=-x^2+5x\) by algebraically determining its key features.
The parabola opens downward.
The \(x\)-intercepts are \((5,0)\) and \((0,0)\text{.}\)
The \(y\)-intercept is \((0,0)\text{.}\)
The vertex is \((2.5,6.25)\) and the axis of symmetry is \(x=2.5\text{.}\)
Graph \(y=-x^2+16\) by algebraically determining its key features.
The parabola opens downward.
The \(x\)-intercepts are \((-4,0)\) and \((4,0)\text{.}\)
The \(y\)-intercept is \((0,16)\text{.}\)
The vertex is \((0,16)\) and the axis of symmetry is \(x=0\text{.}\)
Graph \(y=x^2+4x+7\) by algebraically determining its key features.
The parabola opens upward.
There are no \(x\)-intercepts.
The \(y\)-intercept is \((0,7)\text{.}\)
The vertex is \((-2,3)\) and the axis of symmetry is \(x=-2\text{.}\)
Symmetry can be used to determine the point \((-4,7)\text{.}\)
Graph \(y=x^2-2x+6\) by algebraically determining its key features.
The parabola opens upward.
There are no \(x\)-intercepts.
The \(y\)-intercept is \((0,6)\text{.}\)
The vertex is \((1,5)\) and the axis of symmetry is \(x=1\text{.}\)
Symmetry can be used to determine the point \((2,6)\text{.}\)
Graph \(y=x^2+2x-5\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-1+\sqrt{6},0)\) and \((-1-\sqrt{6},0)\text{,}\) or approximately \((-3.449,0)\) and \((1.449,0)\text{.}\)
The \(y\)-intercept is \((0,-5)\text{.}\)
The vertex is \((-1,-6)\) and the axis of symmetry is \(x=-1\text{.}\)
Symmetry can be used to determine the point \((-2,-5)\text{.}\)
Graph \(y=x^2-6x+2\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((3-\sqrt{7},0)\) and \((3+\sqrt{7},0)\text{,}\) or approximately \((0.354,0)\) and \((5.646,0)\text{.}\)
The \(y\)-intercept is \((0,2)\text{.}\)
The vertex is \((3,-7)\) and the axis of symmetry is \(x=3\text{.}\)
Symmetry can be used to determine the point \((6,2)\text{.}\)
Graph \(y=-x^2+4x-1\) by algebraically determining its key features.
The parabola opens downward.
The \(x\)-intercepts are \((2-\sqrt{3},0)\) and \((2+\sqrt{3},0)\text{,}\) or approximately \((0.268,0)\) and \((3.732,0)\text{.}\)
The \(y\)-intercept is \((0,-1)\text{.}\)
The vertex is \((2,3)\) and the axis of symmetry is \(x=2\text{.}\)
Symmetry can be used to determine the point \((4,-1)\text{.}\)
Graph \(y=-x^2-x+3\) by algebraically determining its key features.
The parabola opens downward.
The \(x\)-intercepts are \(\left(\frac{-1-\sqrt{13}}{2},0\right)\) and \(\left(\frac{-1+\sqrt{13}}{2},0\right)\text{,}\) or approximately \((-2.303,0)\) and \((1.303,0)\text{.}\)
The \(y\)-intercept is \((0,3)\text{.}\)
The vertex is \((-0.5,3.25)\) and the axis of symmetry is \(x=-0.5\text{.}\)
Symmetry can be used to determine the point \((-1,3)\text{.}\)
Graph \(y=2x^2-4x-30\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-3,0)\) and \((5,0)\text{.}\)
The \(y\)-intercept is \((0,-30)\text{.}\)
The vertex is \((1,-32)\) and the axis of symmetry is \(x=1\text{.}\)
Symmetry can be used to determine the point \((2,-30)\text{.}\)
Graph \(y=3x^2+21x+36\) by algebraically determining its key features.
The parabola opens upward.
The \(x\)-intercepts are \((-4,0)\) and \((-3,0)\text{.}\)
The \(y\)-intercept is \((0,36)\text{.}\)
The vertex is \((-3.5,-0.75)\) and the axis of symmetry is \(x=-3.5\text{.}\)
Symmetry can be used to determine the point \((-7,36)\text{.}\)
Applications of Quadratic Functions
