Find 2 Quadratic Equations from Coordinates + Shared Maximum

[devitius]

Find two quadratic functions f and g such that f(1) = 0, g(1) = 0, f(0) = 10, g(0) = 10 and which have the maximum value of 18.

My Working:

f(x) = ax[sup:n0ph29d9]2[/sup:n0ph29d9] + bx + c

f(0) = 10
10 = a(0)[sup:n0ph29d9]2[/sup:n0ph29d9] + b(0) + c
c = 10 ---> f(x) = ax[sup:n0ph29d9]2[/sup:n0ph29d9] + bx + 10

f(1) = 0
0 = a(1)[sup:n0ph29d9]2[/sup:n0ph29d9] + b(1) + 10
0 = a + b + 10
a + b = -10 (eq. 1)


g(x) = ax[sup:n0ph29d9]2[/sup:n0ph29d9] + bx + c

g(x) has same coordinates as f(x), so;
g(x) = f(x) = ax[sup:n0ph29d9]2[/sup:n0ph29d9] + bx + 10

Because the question is defining both of these quadratics as having a maximum, they must be negative parabolas... a < 0

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That's about it... can't seem to see the way this is suppose to be solved.

 

[galactus]

The idea is to find two different parabolas, it would appear. You used a,b, and c for both. That implies they have the same coefficients.

The max value of 18 is the y value of the parabolas when the derivatives are 0.

 

[devitius]

Good point - g(x) should instead be defined as dx[sup:2skvi7zm]2[/sup:2skvi7zm] + ex + 10 to specify that it is a quadratic unique to f(x)

Still - doing so doesn't seem to help. I still don't know where to go from here?

Also, I suspect this question was suppose to be solved without the use of differentiation, as I stumbled across it in a section on "Functions & Relations" directed to students who have not yet covered differentation.

 

[galactus]

The vertex of a parabola can be found without differentiating by using \(\displaystyle x=\frac{-b}{2a}\).

 

[devitius]

Sure - that's a part of the quadratic equation.

But I still can't see how that would help for this question - I still have unknowns in a and b and no more than one derived equation in a + b = -10?

 

[galactus]

Yes, there is enough info.

By using the x=-b/(2a), plug into the equation:

\(\displaystyle a(\frac{-b}{2a})^{2}+b(\frac{-b}{2a})+10=18\)

Which simplifies to:

\(\displaystyle 10-\frac{b^{2}}{4a}=18\)

\(\displaystyle a=\frac{-b^{2}}{32}\)

Sub into \(\displaystyle a+b+10=0\), which comes from f(1)=0

This gives one variable, b.

\(\displaystyle \frac{-b^{2}}{32}+b+10=0\)

\(\displaystyle b=40, \;\ b=-8\)

There are the two b's. Now find the two a's and you have it.

After looking at it more closely, we do not need to concern ourselves with e,f,g in the other quadratic because it takes care of itself.

 

[BigGlenntheHeavy]

\(\displaystyle No \ calculus \ necessary.\)

\(\displaystyle f(x) \ = \ ax^2+bx+c\)

\(\displaystyle f(0) \ = \ 10 \ = \ c\)

\(\displaystyle f(1) \ = \ 0 \ = \ a+b+c \ = \ a+b+10\)

\(\displaystyle f(-\frac{b}{2a}) \ = \ 18 \ \implies \ a \ = \ -\frac{b^2}{32}\)

\(\displaystyle Hence, \ -\frac{b^2}{32}+b+10 \ = \ 0 \ \implies \ b \ = \ 40 \ or \ b \ = \ -8\)

\(\displaystyle Ergo, \ f(x) \ = \ -50x^2+40x+10 \ and \ g(x) \ = \ -2x^2-8x+10, \ see \ galactus's \ graph \ above\)

 

[galactus]

Isn't that what I just done?.

 

[BigGlenntheHeavy]

\(\displaystyle Yes \ it \ is \ galactus, \ but \ it \ never \ hurts \ to \ get \ a \ second \ opinion, \ and\)

\(\displaystyle if \ the \ second \ opinion \ is \ the \ same \ as \ the \ first, \ then \ we \ have \ re-inforcement,\)

\(\displaystyle which \ can \ never \ hurt \ the \ aspirant, \ as \ when \ everything \ else \ fails, \ always \ re-inforce\)

\(\displaystyle success \ by \ success \ (when \ you \ knock \ a \ man \ down, \ make \ sure \ he \ stays \ down, \ otherwise \ he's\)

\(\displaystyle \ liable \ to \ get \ up \ and \ knock \ you \ down \ permanently). \ In \ fact, \ what \ this \ country \ needs, \ is \ more\)

\(\displaystyle \ re-inforcement.\)

 

[galactus]

Okey-doke