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Graphing Quadratic Equations Using the Axis of Symmetry

A quadratic equation is a polynomial equation of degree $2$ . The standard form of a quadratic equation is

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

where $a, b$ and $c$ are all real numbers and $a \neq 0$ .

If we replace $0$ with $y$ , then we get a quadratic function

$y = a x^{2} + b x + c$

whose graph will be a parabola .

The axis of symmetry of this parabola will be the line $x = - \frac{b}{2 a}$ . The axis of symmetry passes through the vertex, and therefore the $x$ -coordinate of the vertex is $- \frac{b}{2 a}$ . Substitute $x = - \frac{b}{2 a}$ in the equation to find the $y$ -coordinate of the vertex. Substitute few more $x$ -values in the equation to get the corresponding $y$ -values and plot the points. Join them and extend the parabola.

Example 1:

Graph the parabola $y = x^{2} - 7 x + 2$ .

Compare the equation with $y = a x^{2} + b x + c$ to find the values of $a$ , $b$ , and $c$ .

Here, $a = 1, b = - 7$ and $c = 2$ .

Use the values of the coefficients to write the equation of axis of symmetry .

The graph of a quadratic equation in the form $y = a x^{2} + b x + c$ has as its axis of symmetry the line $x = - \frac{b}{2 a}$ . So, the equation of the axis of symmetry of the given parabola is $x = \frac{- (- 7)}{2 (1)}$ or $x = \frac{7}{2}$ .

Substitute $x = \frac{7}{2}$ in the equation to find the $y$ -coordinate of the vertex.

$\begin{array}{l} y = {(\frac{7}{2})}^{2} - 7 (\frac{7}{2}) + 2 \\ = \frac{49}{4} - \frac{49}{2} + 2 \\ = \frac{49 - 98 + 8}{4} \\ = - \frac{41}{4} \end{array}$

Therefore, the coordinates of the vertex are $(\frac{7}{2}, - \frac{41}{4})$ .

Now, substitute a few more $x$ -values in the equation to get the corresponding $y$ -values.

$x$	$y = x^{2} - 7 x + 2$
$0$	$2$
$1$	$- 4$
$2$	$- 8$
$3$	$- 10$
$5$	$- 8$
$7$	$2$

Plot the points and join them to get the parabola.

Math diagram

Example 2:

Graph the parabola $y = - 2 x^{2} + 5 x - 1$ .

Compare the equation with $y = a x^{2} + b x + c$ to find the values of $a$ , $b$ , and $c$ .

Here, $a = - 2, b = 5$ and $c = - 1$ .

Use the values of the coefficients to write the equation of axis of symmetry.

Substitute $x = \frac{5}{4}$ in the equation to find the $y$ -coordinate of the vertex.

$\begin{array}{l} y = - 2 {(\frac{5}{4})}^{2} + 5 (\frac{5}{4}) - 1 \\ = \frac{- 50}{16} + \frac{25}{4} - 1 \\ = \frac{- 50 + 100 - 16}{16} \\ = \frac{34}{16} \\ = \frac{17}{8} \end{array}$

Therefore, the coordinates of the vertex are $(\frac{5}{4}, \frac{17}{8})$ .

Now, substitute a few more $x$ -values in the equation to get the corresponding $y$ -values.

$x$	$y = - 2 x^{2} + 5 x - 1$
$- 1$	$- 8$
$0$	$- 1$
$1$	$2$
$2$	$1$
$3$	$- 4$

Plot the points and join them to get the parabola.

Math diagram

Example 3:

Graph the parabola $x = y^{2} + 4 y + 2$ .

Here, $x$ is a function of $y$ . The parabola opens "sideways" and the axis of symmetry of the parabola is horizontal. The standard form of equation of a horizontal parabola is $x = a y^{2} + b y + c$ where $a$ , $b$ , and $c$ are all real numbers and $a \neq 0$ and the equation of the axis of symmetry is $y = \frac{- b}{2 a}$ .

Compare the equation with $x = a y^{2} + b y + c$ to find the values of $a$ , $b$ , and $c$ .

Here, $a = 1, b = 4$ and $c = 2$ .

Use the values of the coefficients to write the equation of axis of symmetry.

The graph of a quadratic equation in the form $x = a y^{2} + b y + c$ has as its axis of symmetry the line $y = - \frac{b}{2 a}$ . So, the equation of the axis of symmetry of the given parabola is $y = \frac{- 4}{2 (1)}$ or $y = - 2$ .

Substitute $y = - 2$ in the equation to find the $x$ -coordinate of the vertex.

$\begin{array}{l} x = {(- 2)}^{2} + 4 (- 2) + 2 \\ = 4 - 8 + 2 \\ = - 2 \end{array}$

Therefore, the coordinates of the vertex are $(- 2, - 2)$ .

Now, substitute a few more $y$ -values in the equation to get the corresponding $x$ -values.

$y$	$x = y^{2} + 4 y + 2$
$- 5$	$7$
$- 4$	$2$
$- 3$	$- 1$
$- 1$	$- 1$
$0$	$2$
$1$	$7$

Plot the points and join them to get the parabola.

Math diagram