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] .

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (-2, -1). Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (2, 7). Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3.

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 $y = a x^{2} + b x + c$ is $x = - \frac{b}{2 a} .$

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

The figure shows the steps to find the axis of symmetry for two parabolas. On the left side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times 1, which simplifies to x equals negative 2. On the right side the standard form of a quadratic equation which is y equals a x squared plus b x plus c is written above the given equation y equals negative x squared plus 4 x plus 3. The axis of symmetry is the equation x equals negative b divided by the quantity two times a. Plugging in the values of a and b from the quadratic equation the formula becomes x equals negative 4 divided by the quantity 2 times -1, which simplifies to x equals 2.

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

The figure shows the steps to find the vertex for two parabolas. On the left side is the given equation y equals x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals -2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of -2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with -2 plugged in for the x value which is y equals -2 squared plus 4 times -2 plus 3. This simplifies to y equals -1. Below this is the statement “vertex is (-2, -1)”. On the right side is the given equation y equals negative x squared plus 4 x plus 3. Below the equation is the statement “axis of symmetry is x equals 2”. Below that is the statement “vertex is” next to the statement is an ordered pair with x-value of 2, the same as the axis of symmetry, and the y-value is blank. Below that the original equation is rewritten. Below the equation is the equation with 2 plugged in for the x value which is y equals negative the quantity 2 squared, plus 4 times 2 plus 3. This simplifies to y equals 7. Below this is the statement “vertex is (2, 7)”.

Axis of symmetry and vertex of a parabola

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

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

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

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

Solution

ⓐ
The axis of symmetry is the line $x = - \frac{b}{2 a}$ .
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 $(1, − 1) .$

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

ⓐ $x = 2$ ⓑ $(2, −7)$

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

ⓐ $x = 1$ ⓑ $(1, −5)$

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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 $0 = a x^{2} + b x + c$ for $x$ .

\begin{matrix} y = a x^{2} + b x + c \\ 0 = a x^{2} + b x + c \end{matrix}

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 $y = x^{2} + 4 x + 3$ .


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

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


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 $(2 + \sqrt{7}, 0)$ and $(2 - \sqrt{7}, 0)$ .

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

\begin{matrix} (2 + \sqrt{7}, 0) \approx (4.6, 0) & (2 - \sqrt{7}, 0) \approx (−0.6, 0) \end{matrix}

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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