Find the x and y Intercepts y=x^2-4x-32

$y = x^{2} - 4 x - 32$

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = x^{2} - 4 x - 32$

Solve the equation.

Rewrite the equation as $x^{2} - 4 x - 32 = 0$ .

$x^{2} - 4 x - 32 = 0$

Factor $x^{2} - 4 x - 32$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 32$ and whose sum is $- 4$ .

$- 8, 4$

Write the factored form using these integers.

$(x - 8) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 8 = 0$

$x + 4 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 8) (x + 4) = 0$ true.

$x = 8, - 4$

x-intercept(s) in point form.

x-intercept(s): $(8, 0), (- 4, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = {(0)}^{2} - 4 \cdot 0 - 32$

Solve the equation.

Remove parentheses.

$y = 0^{2} - 4 \cdot 0 - 32$

Remove parentheses.

$y = {(0)}^{2} - 4 \cdot 0 - 32$

Simplify ${(0)}^{2} - 4 \cdot 0 - 32$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 0 - 4 \cdot 0 - 32$

Multiply $- 4$ by $0$ .

$y = 0 + 0 - 32$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 32$

Subtract $32$ from $0$ .

$y = - 32$

y-intercept(s) in point form.

y-intercept(s): $(0, - 32)$

Step 3

List the intersections.

x-intercept(s): $(8, 0), (- 4, 0)$

y-intercept(s): $(0, - 32)$

Step 4

Find the x and y Intercepts y=x^2-4x-32

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