Answer
Opens upward
Vertex: $(−1,−3)$
x-intercepts: $(−2.22,0)$, $(.22,0)$
y-intercept: $(-1,-3)$
Work Step by Step
$f(x)=2x^2+4x-1$
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 $x=−b/2a$.
$a=2$, $b=4$, $c=-1$
$x=−4/2*2$
$x=−4/4$
$x=−1$
$x=−1$
$f(x)=2x^2+4x-1$
$f(−1)=(−1)^2+4*−1-1$
$f(−1)=1−4-1$
$f(−1)=−4$
If we let $x=0$, we can find the y-intercept.
$x=0$
$f(x)=2x^2+4x-1$
$f(-1)=2*(-1)^2+4*-1-1$
$f(-1)=2*1-4-1$
$f(-1)=2-4-1$
$f(-1)=-3$
If we let $f(x)$=0, we can find the x-intercepts.
$f(x)=2x^2+4x-1$
$0=2x^2+4x-1$
$a=2$, $b=4$, $c=-1$
$x=(−b±\sqrt{b^2−4ac})/2a$
$x=(−4±\sqrt{4^2−4*2*-1})/2*2$
$x=(-4±\sqrt{16+8})/4$
$x=(-4±\sqrt{24})/4$
$x=(-4±4.899)/4$
$x=(-4-4.899)/4$
$x=(-8.899)/4$
$x=-2.224$
$x=-2.22$
$x=(-4+4.899)/4$
$x=(.899)/4$
$x=.224$
$x=.22$
