Question

CAN ANYBODY HELP ME WITH THIS ONE TOO!!!!

Answer 1

Answer: K

Step-by-step explanation:

Compare the equations of:  h = -t² + 4t + 5  and  h = -t² + 4t + 12

h = -t² + 4t + 5:

• c-value is 5
• ⇒ h-intercept is 5

• intercept form: h = -(t + 1)(t - 5)
• ⇒ intercepts are t = -1 and t = 5

• vertex form: h = -(x - 2)² + 9
• ⇒ vertex is (2, 9)
• ⇒ maximum value of h is 9

h = -t² + 4t + 12:

• c-value is 12
• ⇒ h-intercept is 12

• intercept form: h = -(t + 2)(t - 6)
• ⇒ intercepts are t = -2 and t = 6

• vertex form: h = -(x - 2)² + 16
• ⇒ vertex is (2, 16)
• ⇒ maximum value of h is 16

The only value that changed between these two equations is "c", however the h-intercepts, maximum values, and t-intercepts are all different.  

This is because the c-value shifted the entire parabola up.  Shifting the parabola up or down will affect the intercepts and the maximum value of the vertex.

Answer 2

Answer:

I only

Step-by-step explanation:

given h = - at² + bt + c

the vertex = - $\frac{b}{2a}$

Thus altering a or b effects the position of the vertex

and the maximum value is the value of the vertex

Altering c effects the h-intercept only