SOLUTION: Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function. f(x) = -x^2 + 4x + 6

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function. f(x) = -x^2 + 4x + 6      Log On


   



Question 131500: Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function.
f(x) = -x^2 + 4x + 6

Answer by solver91311(24713) About Me  (Show Source):
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The x-coordinate of the vertex of a parabola in the form f%28x%29=ax%5E2%2Bbx%2Bc is given by -b%2F2a.

The y-coordinate is then f%28-b%2F2a%29.

The line of symmetry passes through the vertex, so the equation is x=-b%2F2a.

The maximum or minimum is the value of the function at -b%2F2a. Whether it is a maximum or minimum depends on whether the parabola opens up or down. If it is concave up (makes a valley rather than a hill), the point is a minimum, otherwise it is a maximum. You can tell which way the parabola opens by the sign on the lead coefficient. if a%3C0, it is concave down, if a%3E0, it is concave up, and, of course, if a=0 you don't have a parabola at all.

Let's look at your specific problem:

f%28x%29+=+-2x%5E2%2B4x%2B6+

First thing to note is that a%3C0, so this is a concave down parabola and the vertex is a maximum.

-b%2F2a=-4%2F%282%28-2%29%29=1, so the x-coordinate of the vertex is 1 and the equation of the line of symmetry is x=1.

The value of the function at -b%2F2a, denoted f%28-b%2F2a%29 for your problem is f%281%29=-2%281%29%5E2%2B4%281%29%2B6=%28-2%29%2B4%2B6=8

So the y-coordinate of the vertex and the maximum value of f is 8