If the function f(x)=2x^2+mx-3 has an axis of symmetry of x=4, what is the value of m?

Answer 1

If x = 4 is axis of symmetry then x-coordinate of vertex of parabola is 4.

Now let change equation to vertex form:

f(x) = 2x² + mx - 3 = 2(x² + m/2 x - 3/2) = 2·[(x + m/4)² - m²/16 - 3/2] = 2(x + m/4)² - m²/8 - 3.

Because m/4 = - 4 and m = - 16. Then equetion of parabola in vertex form is f(x) = 2(x - 4)² - (- 16)²/8 - 3 or

f(x) = 2(x - 4)² - 35.

Answer m = - 16