quick quiz\n1. graph the following function. label intercepts and the vertex.\n$f(x)=-x^{2}-3x + 10$\n2…

Explanation:

Step1: Find the x - intercepts

Set $f(x)=0$, so $-x^{2}-3x + 10=0$. Multiply through by - 1 to get $x^{2}+3x - 10=0$. Factor: $(x + 5)(x - 2)=0$. Then $x=-5$ or $x = 2$. The x - intercepts are $(-5,0)$ and $(2,0)$.

Step2: Find the y - intercept

Set $x = 0$, then $f(0)=-0^{2}-3\times0 + 10=10$. The y - intercept is $(0,10)$.

Step3: Find the vertex

The x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. For $y=-x^{2}-3x + 10$, $a=-1$, $b=-3$, so $x=-\frac{-3}{2\times(-1)}=-\frac{3}{2}$. Substitute $x = -\frac{3}{2}$ into $y=-x^{2}-3x + 10$: $y=-\left(-\frac{3}{2}\right)^{2}-3\times\left(-\frac{3}{2}\right)+10=-\frac{9}{4}+\frac{9}{2}+10=-\frac{9}{4}+\frac{18}{4}+10=\frac{-9 + 18}{4}+10=\frac{9}{4}+10=\frac{9+40}{4}=\frac{49}{4}$. The vertex is $\left(-\frac{3}{2},\frac{49}{4}\right)$.

Step4: Find where the tangent has slope 0

The derivative of $y=-x^{2}-3x + 10$ is $y^\prime=-2x-3$. Set $y^\prime = 0$, so $-2x-3=0$. Add 3 to both sides: $-2x=3$. Divide by - 2: $x=-\frac{3}{2}$.

Answer:

1. x - intercepts: $(-5,0)$ and $(2,0)$; y - intercept: $(0,10)$; vertex: $\left(-\frac{3}{2},\frac{49}{4}\right)$.
2. $x = -\frac{3}{2}$