equation of a graph...?

Question

There's a parabola - the one where its -x^2... because its a 'n' shape sort of graph
I'm given the maximum point as (1,5)
and i need to put into the form ax^2+bx+3 , so as to find a and b.
this is all the info i am given...how am i supposed to work this out?

Answer 1

f(x) = ax^2 + bx + 3

We know that this parabola goes through (1, 5), therefore,
f(1) = 5. But
f(1) = a(1)^2 + b(1) + 3 = 5, so
a + b + 3 = 5
a + b = 2

We're also given that (1, 5) is a maximum point. That means f'(1) = 0. But

f'(x) = 2ax + b, so
f'(1) = 2a(1) + b
f'(1) = 2a + b = 0

Two equations, two unknowns:

a + b = 2
2a + b = 0

Solving this by whatever means we choose, I'll omit the details and tell you that the solution is a = -2, b = 4.

Therefore, the equation of your parabola is

f(x) = -2x^2 + 4x + 3

Answer 2

graph is -x^2...

let say y =ax^2 + bx +3
point (1,5) is at graph
so, 5 = a(1)^2 + b(1) +3
5 = a + b +3
a+b = 2 --- (1)


dy/dx = 2ax + b

maximum point is (1,5)
so, dy/dx = 2a(1) +b = 0
2a + b = 0----(2)

(2) -(1) a = -2,
b = 4,

so graph is y =-2x^2 + 4x +3

Answer 3

a(1)^2 + b(1) + 3 = 5
a + b = 2
Now, since you know you have a max @1,5, differentiate the function and set the derivative =0 when x = 1:
2ax + b = 0
2a + b = 0
a + b = 2
Subtracting ,
a = -2
-2 + b = 2
b = 4
y = -2x^2 + 4x + 3

Checking,
-2(1) + 4(1) + 3 = 5, and
2(-2)(1) + 4 = 0

Answer 4

y=ax^2+bx+3

In the maximum point dy/dx=0
dy/dx=2*a*x+b
dy/dx(x=1)=2*a+b=0
b=-2a

y(x=1)=a+b+3=5
a+b=2
a+(-2a)=2
a=-2
b=4

y=-2x^2+4x+3