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

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Answer 1 · 2018-04-20T06:00:14

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

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

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

$rArr x^2+2x-3=0$

Split the middle term to find the factors.

$rArr x^2+3x-1x-3=0$

$rArr x(x+3)-1(x+3)=0$

$rArr (x+3)(x-1)=0$

$x+3 = 0 or x-1 =0$

$x=-3 or x=1$

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

$color(green)("Step 2"$

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

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

$rArr y = (0)^2+2(0)-3$

$:. y = -3$

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

You can verify these solutions by analyzing the quadratic graph:

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