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

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Answer 1 · 2017-04-16T21:31:47

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

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

Doing so,

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

$y = 4 (- 5)$ $y=4(-5)$

$y = - 20$ $y=-20$

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

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

Doing so,

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

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

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

$x_{1} = - 6.899$ $x_1=-6.899$ and $x_{2} = 2.899$ $x_2=2.899$

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

What are the intercepts of y=4(x−5)+x2y = 4(x - 5)+x^2?