# 10.5 Graphing quadratic equations  (Page 2/15)

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We show the same two graphs again with the axis of symmetry in red. See [link] .

The equation of the axis of symmetry    can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of $y=a{x}^{2}+bx+c$ is $x=-\frac{b}{2a}.$

So, to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $x=-\frac{b}{2a}$ .

Look back at [link] . Are these the equations of the dashed red lines?

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. This point is called the vertex    of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its x -coordinate is $-\frac{b}{2a}$ . To find the y -coordinate of the vertex, we substitute the value of the x -coordinate into the quadratic equation.

## Axis of symmetry and vertex of a parabola

For a parabola with equation $y=a{x}^{2}+bx+c$ :

• The axis of symmetry of a parabola is the line $x=-\frac{b}{2a}$ .
• The vertex is on the axis of symmetry, so its x -coordinate is $-\frac{b}{2a}$ .

To find the y -coordinate of the vertex, we substitute $x=-\frac{b}{2a}$ into the quadratic equation.

For the parabola $y=3{x}^{2}-6x+2$ find: the axis of symmetry and the vertex.

## Solution

 ⓐ The axis of symmetry is the line $x=-\frac{b}{2a}$ . Substitute the values of a, b into the equation. Simplify. $x=1$ The axis of symmetry is the line $x=1$ . ⓑ The vertex is on the line of symmetry, so its x -coordinate will be $x=1$ . Substitute $x=1$ into the equation and solve for y. Simplify. This is the y -coordinate. $y=-1$ The vertex is $\left(1,\text{−}1\right).$

For the parabola $y=2{x}^{2}-8x+1$ find: the axis of symmetry and the vertex.

$x=2$ $\left(2,-7\right)$

For the parabola $y=2{x}^{2}-4x-3$ find: the axis of symmetry and the vertex.

$x=1$ $\left(1,-5\right)$

## Find the intercepts of a parabola

When we graphed linear equations, we often used the x - and y -intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y -intercept    the value of $x$ is zero. So, to find the y -intercept, we substitute $x=0$ into the equation.

Let’s find the y -intercepts of the two parabolas shown in the figure below.

At an x -intercept , the value of $y$ is zero. To find an x -intercept, we substitute $y=0$ into the equation. In other words, we will need to solve the equation $0=a{x}^{2}+bx+c$ for $x$ .

$\begin{array}{}\\ \\ y=a{x}^{2}+bx+c\hfill \\ 0=a{x}^{2}+bx+c\hfill \end{array}$

But solving quadratic equations like this is exactly what we have done earlier in this chapter.

We can now find the x -intercepts of the two parabolas shown in [link] .

First, we will find the x -intercepts of a parabola    with equation $y={x}^{2}+4x+3$ .

 Let $y=0$ . Factor. Use the zero product property. Solve. The x intercepts are $\left(\text{−}1,0\right)$ and $\left(\text{−}3,0\right).$

Now, we will find the x -intercepts of the parabola with equation $y=-{x}^{2}+4x+3$ .

 Let $y=0$ . This quadratic does not factor, so we use the Quadratic Formula. $a=-1$ , $b=4$ , $c=3$ Simplify. The x intercepts are $\left(2+\sqrt{7},0\right)$ and $\left(2-\sqrt{7},0\right)$ .

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.

$\begin{array}{cccc}\left(2+\sqrt{7},0\right)\approx \left(4.6,0\right)\hfill & & & \left(2-\sqrt{7},0\right)\approx \left(-0.6,0\right)\hfill \end{array}$