Find the x and y Intercepts y=x^2+3x-4

$y = x^{2} + 3 x - 4$

Find the x-intercepts.

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

$0 = x^{2} + 3 x - 4$

Solve the equation.

Rewrite the equation as $x^{2} + 3 x - 4 = 0$ .

$x^{2} + 3 x - 4 = 0$

Factor $x^{2} + 3 x - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$(x - 1) (x + 4) = 0$

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

$x - 1 = 0$

$x + 4 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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 $(x - 1) (x + 4) = 0$ true.

$x = 1, - 4$

x-intercept(s) in point form.

x-intercept(s): $(1, 0), (- 4, 0)$

Find the y-intercepts.

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

$y = {(0)}^{2} + 3 (0) - 4$

Solve the equation.

Remove parentheses.

$y = 0^{2} + 3 (0) - 4$

Remove parentheses.

$y = {(0)}^{2} + 3 (0) - 4$

Simplify ${(0)}^{2} + 3 (0) - 4$ .

Simplify each term.

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

$y = 0 + 3 (0) - 4$

Multiply $3$ by $0$ .

$y = 0 + 0 - 4$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 4$

Subtract $4$ from $0$ .

$y = - 4$

y-intercept(s) in point form.

y-intercept(s): $(0, - 4)$

List the intersections.

x-intercept(s): $(1, 0), (- 4, 0)$

y-intercept(s): $(0, - 4)$

Find the x and y Intercepts y=x^2+3x-4

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