Question

Plsssss helpppp teacher won’t helppp

Answer 1

Answer:

A. The graphs of two of the functions have a minimum point.

Step-by-step explanation:

Option A is false, because the only function that has a minimum point is

$h(x)=\frac{1}{4}x^2+1$

Option B is true because all the functions are of the form;

$y=ax^2+c$

The equation of axis of symmetry of equations in this form is x=0.

Option C is also true because, $h(x)=\frac{1}{4}x^2+1$ is a minimum graph and its y-intercept is 1. This graph will hang above the x-axis.

$g(x)=-2x^2-5$ is a maximum graph whose y-intercept is -5.

This graph also hangs below the x-axis.

Option D is also true. The y-intercepts are 6,-5, and 1