Math  /  Algebra

QuestionFind the quadratic function f(x)=ax2+bx+cf(x)=a x^{2}+b x+c for which f(1)=1,f(3)=9f(1)=1, f(-3)=9, and f(3)=33f(3)=33.
What is the function? f(x)=f(x)= \square (Simplify your answer.)

Studdy Solution

STEP 1

What is this asking? We need to find the *coefficients* aa, bb, and cc of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c given three points it passes through: (1,1)(1, 1), (3,9)(-3, 9), and (3,33)(3, 33). Watch out! Don't mix up the xx and yy values when plugging the points into the function!
Also, be careful with your signs when working with negative numbers.

STEP 2

1. Set up equations
2. Solve for *a*
3. Solve for *b*
4. Solve for *c*
5. Write the function

STEP 3

Alright, let's **plug in** our given points into our function!
For the point (1,1)(1, 1), we have f(1)=1f(1) = 1, so a(1)2+b(1)+c=1a(1)^2 + b(1) + c = 1, which simplifies to a+b+c=1a + b + c = 1.

STEP 4

Next, for point (3,9)(-3, 9), we have f(3)=9f(-3) = 9, so a(3)2+b(3)+c=9a(-3)^2 + b(-3) + c = 9.
This simplifies to 9a3b+c=99a - 3b + c = 9.

STEP 5

Finally, for (3,33)(3, 33), we have f(3)=33f(3) = 33, so a(3)2+b(3)+c=33a(3)^2 + b(3) + c = 33.
This simplifies to 9a+3b+c=339a + 3b + c = 33.

STEP 6

Now, we have a system of three equations with three unknowns!
Let's **solve** for aa first.
Notice something cool: the first and third equations have the same coefficient for cc.
Let's subtract the first equation from the third equation: (9a+3b+c)(a+b+c)=331(9a + 3b + c) - (a + b + c) = 33 - 1.
This simplifies to 8a+2b=328a + 2b = 32.

STEP 7

We can simplify this further by dividing both sides by 22, giving us 4a+b=164a + b = 16.

STEP 8

Now, let's subtract the second equation from the third equation: (9a+3b+c)(9a3b+c)=339(9a + 3b + c) - (9a - 3b + c) = 33 - 9.
This simplifies to 6b=246b = 24, so b=4b = 4!

STEP 9

Substitute b=4b = 4 back into 4a+b=164a + b = 16: 4a+4=164a + 4 = 16.
Subtracting 44 from both sides gives 4a=124a = 12, so a=3a = 3!

STEP 10

Hey, we already found b=4b = 4 in the previous step!
Sometimes, things just work out like that!

STEP 11

Let's use the first equation, a+b+c=1a + b + c = 1, and substitute the values we found for aa and bb: 3+4+c=13 + 4 + c = 1.
This simplifies to 7+c=17 + c = 1.
Subtracting 77 from both sides gives us c=6c = -6.

STEP 12

We found a=3a = 3, b=4b = 4, and c=6c = -6.
So, our quadratic function is f(x)=3x2+4x6f(x) = 3x^2 + 4x - 6.

STEP 13

f(x)=3x2+4x6f(x) = 3x^2 + 4x - 6

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