Question

Can someone explain?

Answer 1

The correct statement regarding the quadratic function is:

K. I, II and III.

What is a quadratic function?

A quadratic function is given according to the following rule:

$y = ax^2 + bx + c$

If a > 0, it has a maximum value, and if a < 0, it has a minimum value.

The extreme value is $(x_v,y_v)$, in which:

• $x_v = -\frac{b}{2a}$

• $y_v = -\frac{\Delta}{4a}$

• $\Delta = b^2 - 4ac$

The solutions are:

• $x_1 = \frac{-b + \sqrt{\Delta}}{2a}$

• $x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

In this problem, we have a function h(t). Changing the coefficient c, the h-intercept h(0) changes. Looking at the formulas in the bullet point, the value of $\Delta$ changes, meaning that both the maximum value $y_v$ and the t-intercepts $t_1$ and $t_2$ will change, so option K is correct.

More can be learned about quadratic functions at brainly.com/question/24737967

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Answer 2

Answer: K

Step-by-step explanation:

I. Correct. When t=0, h=c, so if you alter the value of c, the h-intercept changes.

II. Correct, the maximum value of $-at^2 +bt$ will stay the same, but you are changing c, which is the amount you are always adding to it, meaning the maximum will change.

III. Correct. If you plug into the quadratic formula setting h=0, it is easy to see changing the value of c will change the solutions.