Find f(x)=ax^2+bx+c with tangent m=10 at (2,7), x-int at x=1

[crzymath]

find a second degree polynomial f(x)=ax[supik1nw31]2[/supik1nw31]+bx+c such that its graph has a tangent line with slope 10 at the point (2,7) and an x-intercept at (1,0)

when i tried to do this problem i used a system of equations but my professor said that was not the way to approach it.

 

[crzymath]

Re: coefficiant with polynomial

someone please give me an idea on how to do this properly

 

[soroban]

Re: coefficiant with polynomial

Hello, crzymath!

Find a second degree polynomial \(\displaystyle f(x)\:=\:ax^2+bx+c\) such that
its graph has a tangent line with slope 10 at the point (2,7) and an x-intercept at (1,0).

When i tried to do this problem i used a system of equations,
but my professor said that was not the way to approach it.

Interesting . . . I'd like to see your professor's approach.
A system of equation is the only obvious way to go . . .

The point (2,7) is on the graph: .\(\displaystyle f(2) = 7 \quad\Rightarrow\quad 4a + 2b + c \:=\:7\)

The point (1,0) is on the graph: .\(\displaystyle f(1) = 0 \quad\Rightarrow\quad a + b + c \:=\:0\)

\(\displaystyle f'(x) \:=\:2ax + b\:\text{ and }\,f'(2) = 10 \quad\Rightarrow\quad 4a + b \:=\:10\)


Solving the system: .\(\displaystyle a \,=\,3,\;b \,=\, \text{-}2,\;c \,=\, \text{-}1\)

Therefore: .\(\displaystyle f(x) \;=\;3x^2 - 2x - 1\)

 

[crzymath]

Re: coefficiant with polynomial

^^thanks-but where did you get the f '(2) from???

 

[crzymath]

Re: coefficiant with polynomial

i know this is gonna sound really dumb-especially comin from a calculus student-but how would you solve a for 3 systems of equations??? last time i did that was in junior high

 

[mmm4444bot]

... where did you get the f '(2) from?

This is given in the original exercise.

Do you know? The derivative is the slope.