1. ## calc 1

find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0).

2. Originally Posted by weezie23
find 2nd degree polynomial f(x) = ax^2+bx+c such that its graph has a tangent line with slope=6 @ point(-2, 7) and x-int @ (-1, 0).
We can use the points we know to find a system of equations

$f(x) = ax^2+bx+c \to f'(x)=2ax+b$

Since we know the derivative represents slope we know when x=-2 that $f'(-2)=6$ so we get

$E_1: \\\ 6=-4a+b$

Now using the point (-2,7) in f(x) we get

$E_2: \\\ 7=4a-2b+c$

and the point (-1,0) we get

$E_3: \\\ 0=a-b+c$

We now have a system of equations in a,b, and c

Multiplying 3 by -1 and adding it to 2 gives

$7=3a-b$ Now if add equation one to this one we get

$13=-a \iff a=-13$

Plugging this in above gives

$7=3(-13)-b \iff b=-46$

Now putting a and b into equation 3 gives

$0=-13-(-46)+c \iff c=-33$

So we get
$f(x)=-13x^2-46x-33$

I hope this helps.

3. Hello, weezie23!

Find a 2nd degree polynomial: $f(x) \:= \:ax^2+bx+c$ such that its graph
has a tangent line with slope = 6 at point(-2, 7) and x-int (-1, 0)

We have: . $f(x) \:=\:ax^2+bx + c$ .and we must find $a,b,c.$

The point (-2,7) is on the graph.
. . $7 \:=\:a(-2)^2 + b(-2) + c\quad\Rightarrow\quad 4a - 2b + c \:=\:7\;\;{\color{blue}[1]}$

The point (-1,0) is on the graph.
. . $0 \:=\:a(-1)^2 + b(-1) + c\quad\Rightarrow\quad a - b + c \:=\:-1\;\;{\color{blue}[2]}$

When $x = -2$, the slope is 6.
We have: . $f'(x) \:=\:2ax + b$
. . $6 \:=\:2a(-2) + b \quad\Rightarrow\quad -4a + b \:=\:6\;\;{\color{blue}[3]}$

Solve the system of equations: . $\begin{array}{cccc}4a - 2b + c &=& 7 & {\color{blue}[1]} \\ a - b + c &=& 0 & {\color{blue}[2]} \\ -4a + b &=& 6 & {\color{blue}[3]} \end{array}$

$\begin{array}{cccc}\text{Subtract {\color{blue}[2]} from {\color{blue}[1]}} & 3x - b &=&7 \\ \text{Add {\color{blue}[3]}} & \text{-}4a + b &=& 6 \end{array}$

And we have: . $-a \:=\:13\quad\Rightarrow\quad \boxed{a \:=\:-13}$

Substitute into [3]: . $-4(-13) + b \:=\:6 \quad\Rightarrow\quad\boxed{ b \:=\:-46}$

Substitute into [2]: . $-13 + 46 + c \:=\:0 \quad\Rightarrow\quad\boxed{ c \:=\:-33}$

Therefore: . ${\color{red}f(x) \;=\;-13x^2 - 46x - 33}$