Question

find a parabola with equation y = ax^2+bx + c that has slope 12 at x = 1, slope - 20 at x=-1, and passes through the point (2,31). 22. -/7 points (a) for what values of x is the function f(x)=|x^2 - 25| differentiable? (enter your answer using interval notation.) find a formula for f. f(x)=\

$$\begin{cases} & \\text{if }|x|> \\\\ & \\text{if }|x|< \\end{cases}$$

(b) sketch the graph of f.

Explanation:

Step1: Differentiate the parabola equation

The derivative of $y = ax^{2}+bx + c$ with respect to $x$ is $y'=2ax + b$.

Step2: Use the slope - information to form equations

Since the slope at $x = 1$ is $12$, we substitute $x = 1$ into $y'$: $2a\times1 + b=12$, which simplifies to $2a + b=12$.
Since the slope at $x=-1$ is $-20$, we substitute $x=-1$ into $y'$: $2a\times(-1)+b=-20$, which simplifies to $-2a + b=-20$.

Step3: Solve the system of equations for $a$ and $b$

Add the two equations $2a + b=12$ and $-2a + b=-20$ together:
$(2a + b)+(-2a + b)=12+( - 20)$
$2b=-8$, so $b=-4$.
Substitute $b = - 4$ into $2a + b=12$: $2a-4 = 12$, then $2a=16$, and $a = 8$.

Step4: Use the point - passing information to find $c$

The parabola passes through the point $(2,31)$. Substitute $x = 2$, $y = 31$, $a = 8$ and $b=-4$ into $y=ax^{2}+bx + c$:
$31=8\times2^{2}+(-4)\times2 + c$
$31=8\times4-8 + c$
$31=32-8 + c$
$31 = 24 + c$, so $c = 7$.

Answer:

$y=8x^{2}-4x + 7$