What are the intercepts of $y = 4(x - 5)+x^2$ ?

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Answer 1 · 2017-04-16T21:31:47

The x-intercepts are the point of $(2.899,0)$ and $(-6.899,0)$ , the y-intercept is $(0,-20)$

For y-intercept(s), let $x=0$

Doing so,

$y=4(0-5)+0^2$

$y=4(-5)$

$y=-20$

Therefore the y-intercept is $(0,-20)$

For x-intercept(s), let $y=0$

Doing so,

$0=4(x-5)+x^2$

$x^2+4x-20=0$

Use the quadratic formula (i'll let you do that),

$x_1=-6.899$ and $x_2=2.899$

Therefore the x-intercepts are the point of $(2.899,0)$ and $(-6.899,0)$ , the y-intercept is $(0,-20)$

What are the intercepts of y = 4(x - 5)+x^2y = 4(x - 5)+x^2?