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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 4$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 3)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 1}$

Add $16$ and $12$ .

$x = \frac{- 4 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{2}$ .

$x = - 2 \pm \sqrt{7}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 1}$

Add $16$ and $12$ .

$x = \frac{- 4 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{2}$ .

$x = - 2 \pm \sqrt{7}$

Change the $\pm$ to $+$ .

$x = - 2 + \sqrt{7}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{- 4 \pm \sqrt{16 + 12}}{2 \cdot 1}$

Add $16$ and $12$ .

$x = \frac{- 4 \pm \sqrt{28}}{2 \cdot 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{- 4 \pm \sqrt{4 (7)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{7}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{7}}{2}$ .

$x = - 2 \pm \sqrt{7}$

Change the $\pm$ to $-$ .

$x = - 2 - \sqrt{7}$

The final answer is the combination of both solutions.

$x = - 2 + \sqrt{7}, - 2 - \sqrt{7}$

x-intercept(s) in point form.

x-intercept(s): $(- 2 + \sqrt{7}, 0), (- 2 - \sqrt{7}, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

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

Multiply $4$ by $0$ .

$y = 0 + 0 - 3$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

Step 3

List the intersections.

x-intercept(s): $(- 2 + \sqrt{7}, 0), (- 2 - \sqrt{7}, 0)$

y-intercept(s): $(0, - 3)$

Step 4

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

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