Question

f(x) = 2x^2 + 12x - 3
Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find the value.

Answer 1

Complete the square.

Answer 2

Well, there's a simpler way. The leading term of the quadratic is positive, so it opens up, meaning it will have a minimum value. The minimum value being the vertex of the parabola. Simply use the formula \(x = -\dfrac{b}{2a}\) to find x. Then insert the value for x in to the quadratic equation to find f(x), which will yield the minimum value.

Answer 3

ok so if the leading term was negative it would have a maximum value? Would you still use the vertex of the parabola and use the formula x = - b/ 2a to find x?

Answer 4

Yes, but the value of x will not be the maximum value. You have to insert the x in to the quad equation to find y coordinate. The y-coordinate will be the max value in that case.

Answer 5

completing the square isn't hard. It's probably the point of the problem.

Answer 6

That's an assumption you're making @wio. You have no way of knowing how the problem is supposed to be solved. Obviously the user is familiar with the vertex method.

Answer 7

Thanks for your help @Hero and @wio.

Answer 8

Completing square:
\[f(x)=ax^2+bx+c \]\[f(x)=a(x^2+\frac{b}{a}x)+c \]\[f(x)=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2-(\frac{b}{2a})^2)+c \]\[f(x)=a[(x+\frac{b}{2a})^2-(\frac{b}{2a})^2]+c \]\[f(x)=a(x+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c \]\[f(x)=a(x+\frac{b}{2a})^2-(\frac{b^2}{4a})+c \]\[f(x)=a(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a}) \]

The vertex is at \((-\frac{b}{2a}, -\frac{b^2-4ac}{4a}) \).
So, what's so-called a "formula" can be obtained by completing square.