# How do you find the x and y intercepts for y = x^2 + 2x - 3?

Apr 20, 2018

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x-intercepts are : color(blue)((1,0), (-3,0)

y-intercept: color(blue)((0,-3)

#### Explanation:

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Standard Form of a quadratic equation is color(blue)(ax^2+bx+c=0

We have a quadratic function color(red)(y=f(x)=x^2+2x-3

color(green)("Step 1"

x-intercepts is a point on the graph where color(blue)(y=0

Solve color(red)(y=x^2+2x-3=0

$\Rightarrow {x}^{2} + 2 x - 3 = 0$

Split the middle term to find the factors.

$\Rightarrow {x}^{2} + 3 x - 1 x - 3 = 0$

$\Rightarrow x \left(x + 3\right) - 1 \left(x + 3\right) = 0$

$\Rightarrow \left(x + 3\right) \left(x - 1\right) = 0$

$x + 3 = 0 \mathmr{and} x - 1 = 0$

$x = - 3 \mathmr{and} x = 1$

Hence, x-intercepts are $\left(1 , 0\right) , \left(- 3 , 0\right)$

color(green)("Step 2"

y-intercepts is the point on the graph where color(blue)(x=0

Solve color(red)(y=x^2+2x-3, "with x"= 0

$\Rightarrow y = {\left(0\right)}^{2} + 2 \left(0\right) - 3$

$\therefore y = - 3$

Hence, y-intercept is at $\left(0 , - 3\right)$

You can verify these solutions by analyzing the quadratic graph: