How do you find the minimum and maximum value of y=-(x-3)^2-4?

1 Answer
Jun 7, 2017

Maximum: (3,-4)
Minimum: N/A

Explanation:

The graph of the equation is a parabola with vertex (3,-4).

When a parabola is in vertex form, y=a(x-h)^2 +k, the vertex (maximum or minimum) is given by the point (h,k). Because a<0, the parabola opens downward, so it must have a maximum. It does not have a minimum point because the parabola extends downward forever to -∞. In any parabola there is either a minimum or a maximum, so your question should be rephrased: "How do you find the minimum or maximum value of y=-(x-3)^2 -4." graph{y=-(x-3)^2 -4 [-262.6, 262.6, -131.3, 131.3]}

One way to find the maximum or vertex is graphically. As depicted in the graph above, the maximum value of f occurs when x=3 and y=-4. The vertex or maximum is (3,-4)

Another way to find the vertex (h,k) is by putting the equation in vertex form: y=a(x-h)^2 +k. Your equation is already in this form, so just solve for h and k. Solve for h by setting x-3=0 and solving for x=h=3. Then, find k = -4. The vertex or maximum is (3,-4)

If your quadratic function was in standard form (ax^2+bx+c=0), the vertex is given by the point (-b/(2a), f(-b/(2a))). For example, if you were asked to find the maximum or vertex of a quadratic function with equation y=-x^2+6x-13, the x-value would be -b/(2a) = -6/(2*-1) = 3). The y-value would be f(3) = -(3)^2 +6(3)-13 = -4). The vertex or maximum is (3,-4)