Finding the constants of a quadratic function

[1141]

Help, please!
I have this question to do for an assignment, it's the only one I don't understand.

the question is:

The quadratic function f(x) = px[sup:3va7gcd7]2[/sup:3va7gcd7]+qx+r has f(0)=35, f(1)=20 and f(2)=11.
Find the values of the constants p, q and r.

Express (fx) in the form a(x+b)[sup:3va7gcd7]2[/sup:3va7gcd7]. Use your answer to find the smallest value of f(x).

For the second part of the question, "Express (fx) in the form a(x+b)[sup:3va7gcd7]2[/sup:3va7gcd7]...etc" I think I will be able to do that. I know how to put an equation into the completed square form and if I understand correctly, the smallest value would mean the minima?
It's the first part I don't quite understand how to do. I don't understand how I would possibly go about solving it.

Thanks!

 

[soroban]

Hello, 1141!

The quadratic function $\displaystyle f(x) \:=\: px^2+qx+r$ has: .$\displaystyle f(0)\,=\,35,\; f(1)\,=\,20,\;f(2)=11$

(a) Find the values of the constants $\displaystyle p, q, r.$

(b) Express $\displaystyle f(x)$ in the form $\displaystyle a(x+b^2)$
. . .Use your answer to find the smallest value of $\displaystyle f(x).$

You must "read" the given statements.


$\displaystyle f(0) \,=\,35\,\text{ means: }\:\text{"When }x = 0,\,\text{ the function equals }35."$

$\displaystyle \text{So we have: }\;p(0^2) + q(0) + r \:=\:35 \quad\Rightarrow\quad r \:=\:35$ .[1]


$\displaystyle \text{Similarly, we have:}$

. . \(\displaystyle f(1) \,=\,20 \quad\Rightarrow\quad \(1^2) + q(1) + r \:=\:20 \quad\Rightarrow\quad p + q + r \:=\:20\) .[2]

. . $\displaystyle f(2) \,=\,11 \quad\Rightarrow\quad p(2^2) + q(2) + r \:=\:11 \quad\Rightarrow\quad 4p + 2q + r \:=\:11$ .[3]


$\displaystyle \text{And we can solve this system of equations.}$

 

[mmm4444bot]

$\displaystyle \text{And we can solve this system of equations.}$

I am somewhat surprised that you did not.

 

[1141]

Hello, 1141!

You must "read" the given statements.


$\displaystyle f(0) \,=\,35\,\text{ means: }\:\text{"When }x = 0,\,\text{ the function equals }35."$

$\displaystyle \text{So we have: }\;p(0^2) + q(0) + r \:=\:35 \quad\Rightarrow\quad r \:=\:35$ .[1]


$\displaystyle \text{Similarly, we have:}$

. . \(\displaystyle f(1) \,=\,20 \quad\Rightarrow\quad \(1^2) + q(1) + r \:=\:20 \quad\Rightarrow\quad p + q + r \:=\:20\) .[2]

. . $\displaystyle f(2) \,=\,11 \quad\Rightarrow\quad p(2^2) + q(2) + r \:=\:11 \quad\Rightarrow\quad 4p + 2q + r \:=\:11$ .[3]


$\displaystyle \text{And we can solve this system of equations.}$

[/size]

Alright, I understand all that.
But I'm still kind of lost.

I sustituted r=35 into the other two equations to get:

p + q = -15
and
4p + 2q = -24

But I'm not even sure if that is even relevant to what they're asking me. And even if it is, I'm not sure how I would use those.
:?:

 

[Aladdin]

After you find p , q and r ~

You write the given quadratic equation in the form of a(x+b)^2 .

Use this to find the where the smallest value occurs .

 

[1141]

After you find p , q and r ~

You write the given quadratic equation in the form of a(x+b)^2 .

Use this to find the where the smallest value occurs .

I understand that, but it's the part of finding p, q and r that I'm stuck on.

 

[mmm4444bot]

p + q = -15
and
4p + 2q = -24

But I'm not even sure if that is even relevant to what they're asking me.

It is known to be most relevant, glasshoppah.

With two equations, we can solve for p and q using the elimination method.

Click here to see a lesson on the substitution method.

Come back, if you need more help.

Cheers