Question

graph the function (h(x)=-2x^{2}-4x + 3) and label the points on the image.

Explanation:

Step1: Find the vertex

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is $x=-\frac{b}{2a}$. For $h(x)=-2x^{2}-4x + 3$, where $a=-2$ and $b = - 4$, then $x=-\frac{-4}{2\times(-2)}=-1$. Substitute $x = - 1$ into the function: $h(-1)=-2\times(-1)^{2}-4\times(-1)+3=-2 + 4+3=5$. So the vertex is $(-1,5)$.

Step2: Find the y - intercept

Set $x = 0$ in the function $h(x)$. Then $h(0)=-2\times0^{2}-4\times0 + 3=3$. So the y - intercept is $(0,3)$.

Step3: Find the x - intercepts

Set $h(x)=0$, so $-2x^{2}-4x + 3=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, with $a=-2$, $b=-4$, $c = 3$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-4)^{2}-4\times(-2)\times3=16 + 24=40$. Then $x=\frac{4\pm\sqrt{40}}{-4}=\frac{4\pm2\sqrt{10}}{-4}=\frac{2\pm\sqrt{10}}{-2}$.

Step4: Choose other points

Let's choose $x=-2$, then $h(-2)=-2\times(-2)^{2}-4\times(-2)+3=-8 + 8+3=3$. So the point is $(-2,3)$. Let $x = 1$, then $h(1)=-2\times1^{2}-4\times1+3=-2-4 + 3=-3$. So the point is $(1,-3)$.

Answer:

The vertex is $(-1,5)$, y - intercept is $(0,3)$, points $(-2,3)$ and $(1,-3)$ can be used to graph the function. The axis of symmetry is $x=-1$.