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

Nov 30, 2014

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

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

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

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

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

by multiplying by $- 1$,

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

by factoring out,

$0 = \left(x + 6\right) \left(x - 3\right) \implies x = - 6 , 3$

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

I hope that this was helpful.