Graphing Quadratics Using Vertex and Intercepts

Question

Step 5 - Using the quadratic equation, find x-intercept and place on the line provided.  y = −x^2 − 4x + 5 If there is more than one x-intercept, separate the numbers by a comma.

Step 5 - QUADRATIC EQUATION - Using the equation below, to find the x-intercept and write it on the line provided.  y = x^2 − 2x - 8 If there is more than one "x" intercept, separate your answers by a comma. The "x-intercept" are also know a s zero pairs

Answer 1

-x^2 -4x +5

=-(x^2 +4x) +5

complete the square,

add & subtract half the x coefficient squared = (-4/2)^2 = 4,

both inside the parentheses and to the constant term

= -(x^2+4x +4) + 5+4

= -(x+2)^2 + 9 = 0 is vertex form with vertex = (-2,9)

a(x-h)^2 +k has its vertex at (h,k), h=-2, k= 9, (h,k) = (-2,9)

(x+2)^2 = 9

x+2 = +/-3

x = -2+/-3

x = -5, 1

the x coordinate of the vertex is the average of the two zeros: -2 = (-5+1)/2 = -4/2 =-2

x^2-2x-8

= x^2-2x +1 - 8-1

= (x-1)^2 -9 = 0

x-1 = +/-3

x = 1+/-3

x = -2, 4

Answer 2

First problem:

y = -x² – 4x + 5

x = [-(-4) ± ((-4)² - 4(-1)(5)) ^½]/2(-1)

x = [4 ± (16 + 20)^½]/(-2)

x = [4 ± (36) ^½]/(-2)

x = (4 ± 6]/(-2)

x = (4 + 6)/(-2) = -5, or

x = (4 - 6)/(-2) = 1

Therefore, x intercepts are:

x = -5, 1

Second problem:

y = -x² – 2x -8

x = [-(-2) ± ((-2)² - 4(1)(-8)) ^½]/2(1)

x = [2 ± (4 + 32)^½]/2

x = [2 ± (36) ^½]/2

x = (2 ± 6]/2

x = (2 + 6)/2 = 4 or

x = (2 - 6)/2 = -2

Therefore, x intercepts are:

x = -2, 4