Find the x-intercepts (if any) for the graph of the quadratic function.

Question

f(x)=(x+1)^2-1.

Answer 1

Given quadratic function f (x) = (x + 1)^2  - 1

Compare it to parabola vertex form y  = a (x - h )^2 +k

y = (x +1)^2  - 1

Vertex ( h ,k ) = (-1,-1)

To find x  intercept substitute y  = 0 in above function.

0 = (x + 1)^2  - 1

(x + 1)^2 = 1

Apply squre root on each side.

x + 1 = ± 1

x + 1 = 1 and x + 1 = -1

Subtract 1 from each side.

x  = 0 and x  = -2

So the x  intercepts are 0,-2.

The points are (0,0) and (-2,0).

Graph

Draw the coodinate plane.

Plot the intercept points and vertex.

Connect the plotted points like a curve.

Then there formed parabola is indicatng f (x ) = (x + 1)^2 - 1.