Question

Graph the function f(x)=x^2+2x-8

what are x intercepts
what are y intercepts
what is maximum or minimum value

Answer 1

Graph attached

• y=x²+2x-8
• y=x²+4x-2x-8
• y=x(x+4)-2(x+4)
• y=(x+4)(x-2)

x intercepts (-4,0) and (2,0)

Y intercept :-

Put x=0

• y=-8

(0,-8)

Vertex is the minimum

• (-1,-9)

Answer 2

The x intercepts are 2, -4

The y intercept is -8

The minimum is -9

Explanation:

f(x)=x^2+2x-8

To find the x intercepts, set equal to zero and factor

0 =x^2+2x-8

0 = (x+4)(x-2)

Using the zero product property

0 = x+4 0 = x-2

x = -4 x = 2

The x intercepts are 2, -4

To find the y intercepts, set x =0 and solve for y

y = 0^2 +2(0) -8

y = -8

The y intercept is -8

Since the coefficient of the x^2 is positive, the parabola opens up so we have a minimum.

The vertex is halfway between the x intercepts

(-4+2)/2 = -2/2 = -1

To find the minimum substitute x= -1 into the equation

f(x)=x^2+2x-8

f(-1) = (-1)^2 +2(-1)-8 = 1-2-8 = -9

The minimum is -9