Unit 8 quadratic equations homework 2 graphing quadratic equations answer key

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 ≠ 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 = − b 2 a . The axis of symmetry passes through the vertex, and therefore the x -coordinate of the vertex is − b 2 a . Substitute x = − 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 = − b 2 a . So, the equation of the axis of symmetry of the given parabola is x = − ( − 7 ) 2 ( 1 ) or x = 7 2 .

Substitute x = 7 2 in the equation to find the y -coordinate of the vertex.

y = ( 7 2 ) 2 − 7 ( 7 2 ) + 2 = 49 4 − 49 2 + 2 = 49 − 98 + 8 4 = − 41 4

Therefore, the coordinates of the vertex are ( 7 2 , − 41 4 ) .

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

Plot the points and join them to get the parabola.

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.

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 = − b 2 a . So, the equation of the axis of symmetry of the given parabola is x = − ( 5 ) 2 ( − 2 ) or x = 5 4 .

Substitute x = 5 4 in the equation to find the y -coordinate of the vertex.

y = − 2 ( 5 4 ) 2 + 5 ( 5 4 ) − 1 = − 50 16 + 25 4 − 1 = − 50 + 100 − 16 16 = 34 16 = 17 8

Therefore, the coordinates of the vertex are ( 5 4 , 17 8 ) .

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

Plot the points and join them to get the parabola.

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 ≠ 0 and the equation of the axis of symmetry is y = − 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 = − b 2 a . So, the equation of the axis of symmetry of the given parabola is y = − 4 2 ( 1 ) or y = − 2 .

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

x = ( − 2 ) 2 + 4 ( − 2 ) + 2 = 4 − 8 + 2 = − 2

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.

Plot the points and join them to get the parabola.