solve for a and b

Question

the function f(x)=x^2+ax+b has a a relative minimum point at (2,-4), solve for a and b

I can find the first part of the equation, but I don't get how to find the equation that you use to actually find the values of a and b

Answer 1

You need to come up with two equations since there are two unknowns. The point (2, -4) is on the graph. That allows you to set up the first equation. Since that point is also a relative minimum, that means the slope of the tangent line at that point is 0, which in turn means the first derivative is equal to 0 when x = 2.

See if you can take it from there. then check your result here:

desmos.com/calculator/hwxuqradfh

Answer 2

You have two pieces of information here.

First, there is a minimum at (x,y)=(2,-4). This means that f(2)=-4. Pluggin this in, we get the equation 4+2a+b=-4.

Second, because there is a minimum at that point, the derivative must be zero there. So f'(x)=2x+a is equal to zero at x=2. Therefore, f'(2)=4+a=0, which immediately gives us that a=-4.

Plugging a=-4 back into the first equation, we get 4+2(-4)+b=-4, which gives b=0.

Answer 3

f(x) = a(x - 2)² - 4, minimum is at vertex

Can you expand. solve and andswer?