Question Find a quadratic function $y = a x^2 + b x + c$ passing through $(4,4)$ , with tangent line slopes -1 at $x=2$ and 11 at $x=8$ .

Studdy Solution

STEP 1

Assumptions1. The quadratic function is in the form $y = ax^ + bx + c$ . The function passes through the point $(4,4)$ 3. The slope of the tangent line at $x=$ is $-1$
4. The slope of the tangent line at $x=8$ is $11$

STEP 2

The derivative of the quadratic function $y = ax^2 + bx + c$ is $y' =2ax + b$ . This derivative gives us the slope of the tangent line at any point $x$ .

STEP 3

We know that the slope of the tangent line at $x=2$ is $-1$ . So we set $y' = -1$ when $x=2$ .
$-1 =2a(2) + b$

STEP 4

We also know that the slope of the tangent line at $x=8$ is $11$ . So we set $y' =11$ when $x=8$ .
$11 =2a(8) + b$

STEP 5

Now we have a system of two equations $\begin{cases} -1 =4a + b \\11 =16a + b \end{cases}$

STEP 6

Subtract the first equation from the second to solve for $a$ .
$11 - (-1) =16a -4a$

STEP 7

olve for $a$ .
$a = \frac{12}{12} =1$

STEP 8

Substitute $a=1$ into the first equation of the system to solve for $b$ .
$-1 =4(1) + b$

STEP 9

olve for $b$ .
$b = - -4 = -5$

STEP 10

Now we know that $a=$ and $b=-5$ . We can substitute these values into the quadratic function $y = ax^2 + bx + c$ .
$y = x^2 -5x + c$

STEP 11

We know that the function passes through the point $(4,4)$ . So we substitute $x=4$ and $y=4$ into the equation to solve for $c$ .
$4 = (4)^ -5(4) + c$

STEP 12

olve for $c$ .
$c =4 -16 +20 =8$

STEP 13

Now we know that $a=$ , $b=-5$ , and $c=8$ . So the quadratic function is $y = x^2 -5x +8$ This is the quadratic function that passes through the point $(,)$ with the given slopes of the tangent lines at $x=2$ and $x=8$ .

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