# What are the intercepts of y = (x – 1)^2 – 16?

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Given y= (x-1)^2 -16

We need to find the intercepts.

First we will find the y- intercept. Then the value of x is 0.

==> y= -15 ==> y-intercept is ( 0. -15)

Now we will find the x-intercept, then y= 0

==> (x-1)^2 -16 = 0

==> (x-1)^2 = 16

==> x -1 = +-4

==> x = 5 , -3

Then x-intercepts are: ( 5, 0) and (-3,0)

**Then the intercepts for x are ( 5, 0) (-3,0), and the intercept for y is (0, -15).**

When the curve of the equation y = (x - 1)^2 - 16 intercepts the x-axis, y = 0

(x - 1)^2 - 16 = 0

=> (x - 1)^2 = 16

=> x - 1 = 4 and x - 1 = -4

=> x = 5 and x = -3

The x-intercepts are (5 , 0) and (-3, 0)

When the curve intercepts the y-axis x= 0

y = (0 - 1)^2 - 16

=> y = -15

The y-intercept is (0, -15)

** For y = (x - 1)^2 - 16****, the x-intercepts are (5 , 0) and (-3, 0), and the y-intercept is (0, -15).**

The curve defined by the equation y = (x - 1)^2 - 16 is a parabola.

When the curve intersects the x-axis, as the equation of the x axis is y = 0, the y coordinate is equal to 0.

Set this condition and solve the equation that is obtained.

(x - 1)^2 - 16 = 0

(x - 1)^2 = 16

Take the square root of both the sides

x + 1 = +- 4

x = 3 and x = -5

The x-intercepts of the curve defined by y = (x - 1)^2 - 16 are x = 3 and x = -5