Formula and minimum value

[dcowboys107]

The graph of a quadratic function f(x) has its vertex at (5,-6) and passes through the point (2,-51). Find a formula for f(x)

 

I got 15(x-5)^2 - 6 Why is this wrong?

 

Find the minimum value of the function

 

(5x-2)(2x+2). I got -49/20 after multiplying out the equation to get

5x^2+3x-2 (simplified). I did -3/10 and put it back into the equation and got it wrong. . .

 

Any ideas?

[Bignose]

The graph of a quadratic function f(x) has its vertex at (5,-6) and passes through the point (2,-51). Find a formula for f(x)

 

I got 15(x-5)^2 - 6 Why is this wrong?

 

 

it is wrong because if you put 2 into the expression for x, it evaluates to 129, not -51. So, it clearly doesn't answer the question.

 

Find the minimum value of the function

 

(5x-2)(2x+2). I got -49/20 after multiplying out the equation to get

5x^2+3x-2 (simplified). I did -3/10 and put it back into the equation and got it wrong. . .

 

Any ideas?

 

What do you mean by "I did -3/10"? You mean you inserted the value of -3/10 for x? Why did you use that number?

 

Also, (5x-2)(2x+2) isn't equal to 5x^2+3x-2. Also also, it isn't an "equation" because there isn't an equals sign. You have an "expression" or maybe a "function".

[dcowboys107]

I got -3/10 by using -b/2a. I put it into the expression and got said number.

 

How do you find the what the value is then?

[Bignose]

Maximums and minimums of f(x)=0 are found by taking the derivative of f(x) and setting it equal to zero and solving for the value(s) of x that solve f'(x)=0.

 

If [math]f(x)=ax^2 + bx +c[/math] then [math]f'(x)=2ax+b[/math] which when you set equal to zero you will get that -b/2a. Your error stems from that fact that [math](5x-2)(2x+2) \ne 5x^2+3x-2[/math].