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
[tex]h(x)=\frac{1}{4}x^2+1[/tex]
Option B is true because all the functions are of the form;
[tex]y=ax^2+c[/tex]
The equation of axis of symmetry of equations in this form is x=0.
Option C is also true because, [tex]h(x)=\frac{1}{4}x^2+1[/tex] is a minimum graph and its y-intercept is 1. This graph will hang above the x-axis.
[tex]g(x)=-2x^2-5[/tex] 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
