Answer
Opens upward
Vertex: $(-5, -10)$
x-intercepts: $(-8.16, 0)$, $(-1.84,0)$
y-intercept: $(0, 15)$
Work Step by Step
$f(x)=x^2+10x+15$
Please see the screenshot of the graph.
The coefficient of $x^2$ is positive, so the graph opens upward.
The vertex of the graph is at $x=-b/2a$.
$a=1$, $b=10$, $c=15$
$x=-10/2*1$
$x=-10/2$
$x=-5$
$x=-5$
$f(x)=x^2+10x+15$
$f(-5)=(-5)^2+10*-5+15$
$f(-5)=25-50+15$
$f(-5)=-10$
If we let $x=0$, we can find the y-intercept.
$x=0$
$f(x)=x^2+10x+15$
$f(0)=0^2+10*0+15$
$f(0)=0+0+15$
$f(0)=15$
If we let $f(x)=0$, we can find the x-intercepts.
$f(x)=x^2+10x+15$
$0=x^2+10x+15$
$a=1$, $b=10$, $c=15$
$x=(-b±\sqrt{b^2-4ac})/2a$
$x=(-10±\sqrt{10^2-4*1*15})/2*1$
$x=(-10±\sqrt{100-60})/2*1$
$x=(-10±\sqrt{40})/2$
$x=(-10±\sqrt{4*10})/2$
$x=(-10±2\sqrt{10})/2$
$x=(-5±\sqrt{10})$
$x=(-5±3.16)$
$x=-5-3.16$
$x=-8.16$
$x=-5+3.16$
$x=-1.84$
