What is the maximum value that the graph of #-3x^2 - 12x + 15#?

Question

Charles Arocha · Accepted Answer

See explanation.

This is a quadratic function with a negative coefficient of #x^2#, so it reaches its maximum value at the vertex of the parabola. Its coordinates can be calculated as: and but you can also calculate #q# by substituting #p# to the function's formula: #p=12/(-6)=-2# #q=f(-2)=-3*(-2)^2-12*(-2)+15# #q=-3*4+12*2+15=-12+24+15=27# Answer: The maximum value is #27# at #x=-2# This can be checked by looking at the graph: graph{-3x^2-12x+15 [-10, 10, -40, 40]}

Christopher Agresta · Answer

undefined To find the maximum value of the quadratic function -3x^2 - 12x + 15, you need to determine the vertex of the parabola represented by the function. The maximum value occurs at the vertex of the parabola.

The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the formula (-b/2a, f(-b/2a)).

For the function -3x^2 - 12x + 15, a = -3 and b = -12.

Substitute these values into the formula to find the x-coordinate of the vertex:
x = -(-12) / (2 * -3) = 2

To find the corresponding y-coordinate, plug this value of x back into the original function:
f(2) = -3(2)^2 - 12(2) + 15 = -12

So, the maximum value of the function is -12.