How do you determine the x and y intercepts of $y = - x^{2} - 3 x + 18$ $y=-x^2-3x+18$ ?

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Answer 1 · 2014-11-30T15:43:38

To find the $y$ $y$ -intercept, we set $x = 0$ $x=0$ ,

$y = - {(0)}^{2} - 3 (0) + 18 = 18$ $y=-(0)^2-3(0)+18=18$

Hence, the $y$ $y$ -intercept is $18$ $18$ .

To find the $x$ $x$ -intercepts, we set $y = 0$ $y=0$ ,

$0 = - x^{2} - 3 x + 18$ $0=-x^2-3x+18$

by multiplying by $- 1$ $-1$ ,

$0 = x^{2} + 3 x - 18$ $0=x^2+3x-18$

by factoring out,

$0 = (x + 6) (x - 3) \Rightarrow x = - 6, 3$ $0=(x+6)(x-3) => x=-6,3$

Hence, the $x$ $x$ -intercepts are $x = - 6, 3$ $x=-6, 3$ .

I hope that this was helpful.

How do you determine the x and y intercepts of y=−x2−3x+18y=-x^2-3x+18 ?