1. ## Product Rule

Determine a quadratic function $f(x) = ax^2 + bx + c$ whose graph passes through the point $(2, 19)$ and that has a horizontal tangent at $(-1, -8)$.

My work:

$y = mx + b$
$19 = -1(2) + b$
$19 + 2 = b$
$21 = b$

$-8 = a(2)^2 + b(2) + 21$
$-8 = 4a + 2b + 21$
$-29 = 4a + 2b$

I don't know what to do now? I'm not even sure I'm on the right track. Please show me the solution step by step?

Textbook Answer: $f(x) = 3x^2 + 6x - 5$

Note: I'm just beginning to learn derivatives...

2. Ooo you're a bit off there .. We'll have to read through this carefully.

It says the function f(x) passes through the point (2, 19). This means that (2,19) is part of the graph, i.e. if you plug in 2 into the equation of the graph f(x) you will get 19: $f(2) = 19 = a(2)^{2} + b(2) + c = {\color{red}4a + 2b + c = 19}$

It also says that the function f(x) has a horizontal tangent. Think about what this means. A horizontal line has a slope = 0 and remember that a derivative essentially represents the slope of a curve. So, we know that at x = -1, f'(x) = 0:
$f'(x) = \underbrace{2ax + b}_{\text{via power rule}}$

Plugging x = -1:
$f'(-1) = 0 = 2a(-1) + b = {\color{blue}-2a + b = 0}$

Also, it gives us another point of the graph (-1, -8) indicating that:
$f(-1) = -8 = a(-1)^{2} + b(-1) + c = {\color{magenta}a - b + c = -8}$

Now, note all the coloured equations. You'll now have to solve this system of equations Luckily, you have the blue equation to simplify your system into one involving only 2 variables

3. Originally Posted by Macleef
Determine a quadratic function $f(x) = ax^2 + bx + c$ whose graph passes through the point $(2, 19)$ and that has a horizontal tangent at $(-1, -8)$.

My work:

$y = mx + b$
$19 = -1(2) + b$
$19 + 2 = b$
$21 = b$

$-8 = a(2)^2 + b(2) + 21$
$-8 = 4a + 2b + 21$
$-29 = 4a + 2b$

I don't know what to do now? I'm not even sure I'm on the right track. Please show me the solution step by step?

Textbook Answer: $f(x) = 3x^2 + 6x - 5$

Note: I'm just beginning to learn derivatives...
$f(x) = ax^2 + bx + c$
$f'(x)=2ax+b$

since there is a horizontal tangent (-1,-8) the derivative is 0.
This gives us our first equation
$0=-2a+b \iff b=2a$

Now using the other point (2,19) and (-1,-8) again

we get two more equations.

$19=4a+2b+c$
$-8=a-b+c$
$b=2a$

subtracting the 2nd from the first we get

$27=3a+3b \iff 9=a+b$

Now subbing the last into the above equation we get

$9=a+2a \iff 9=3a \iff a=3$

This gives b= 6 and c=-5

Now we get

$f(x)=3x^2+6x-5$

edit: Wow I am really late.
I guess this is what I get for working while cooking dinner

4. I wish I knew how to cook o.O ...