There are three variables and only two equations?

The key is that f(1) is a minimum. That means the vertex will be at (1,5). It also means the graph is symmetric on either side of x=1. In other words, if (2,7) is a solution then (0.7) will also be a solution.

Now you have 3 points and can solve it:

Plug in the point (0,7) --> f(0) = 7

7 = a(0)² + b(0) + c

7 = 0 + 0 + c

c = 7

Plug in the point (1,5) --> f(1) = 5

5 = a(1)² + b(1) + 7

a + b = -2

Plug in the point (2,7) --> f(2) = 7

7 = a(2)² + 2b + 7

4a + 2b = 0

Double the first equation and subtract from the second:

2(a + b) = -4

2a + 2b = -4

Subtract:

4a + 2b = 0

2a + 2b = -4

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2a = 0 - (-4)

2a = 4

a = 4/2

a = 2

Finally solve for b:

a + b = -2

2 + b = -2

b = -4

Answer:

f(x) = 2x² - 4x + 7

a = 2

b = -4

c = 7

Double-check with the graph below

There are three pieces of information to fix a, b, and c. These do not have to be equations.

f(x) = ax² + bx + c

The x-coordinate of the vertex of f is -b/(2a) = 1 and f(1) = 5

5 = a + b + c

7 = 4a + 2b + c

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2 = 3a + b

Now, -b/(2a) = 1 ⇒ b = -2a

2 = 3a - 2a = a ☜

b = -4 ☜

5 = a + b + c = 2 - 4 + c = -2 + c ⇒ c = 7 (which we can also get from symmetry)

f(x) = 2x² - 4x + 7 ☜ ☜

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

dy/dx=2ax+b

dy/dx=0 when x=1 so

2a+b=0 and we are given two points. (1,5) and (2,7) so let's set them up

a+b+c=5 and 4a+2b+c=7 getting the difference yields:

3a+b=2 and we know that 2a+b=0, and getting the difference of these two gives:

a=2, using this value of a in 3a+b=2 gives:

6+b=2, so b=-4, using a=2, b=-4 in a+b+c=5

2-4+c=5, so c=7

so a=2, b=-4, and c=7

so the equation is:

f(x)=2x^2-4x+7

f'(x) = 2ax + b

At the minimum, f'(x) = 0, so f'(1) = 0, so we have:

2*a*1 + b = 0

2a + b = 0

f(1) = 5, so we have a + b + c = 5

f(2) = 7, so we have 4a + 2b + c = 7

Now we have 3 equations, and we can solve them to get:

a = 2, b = -4, c = 7

Actually, there are three equations.

5 = a + b + c

7 = 4a + 2b + c

0 = 2a + b <== this equation is true because the slope of f(x) is zero at the minimum, and the slope is f'(x) = 2ax + b.

f(x)=ax^2+bx+c=>

f '(x)=2ax+b=>

f"(x)=2a

f '(x)=0=>2ax+b=0=>

x= -b/2a

f(1)=5 is a min.=>

a+b+c=5----------(1)

2a= -b-------------(2)

f(2)=4a+2b+c=7-------(3)

So, you actually has 3 equations for

3 variables.

Solving the system of (1), (2), (3) get

a=2

b=-4

c=7