Question

Compare f(x) = x² + 6x – 4 to a quadratic function g(x) that has a y-intercept of -6 and the vertex is (-1, -9).

Which function has a lesser minimum?

Answer 1

The function g(x) = -3(x+1)² - 9 has a lesser minimum than f(x) = x² + 6x - 4.

Step-by-step explanation:

A quadratic function is of the form f(x) = ax² + bx + c, where a, b, and c are constants. The vertex form of a quadratic function is given by g(x) = a(x-h)² + k, where (h, k) is the vertex of the parabola.

Comparing the given functions:

• f(x) = x² + 6x - 4
• g(x) = a(x+1)² - 9

To compare the minimum values of the two functions, we need to determine the value of a in g(x). Using the fact that the y-intercept of g(x) is -6, we can substitute x=0 and y=-6 in the equation g(x) = a(x+1)² - 9 to solve for a.

Substituting x=0 and y=-6, we get -6 = a(0+1)² - 9. Solving this equation, we find a = -3.

Therefore, the function g(x) = -3(x+1)² - 9 has a lesser minimum than f(x) = x² + 6x - 4.