Question

Plzz Help ASAP!! i posted the picture?

Answer 1

You can write the function as

h = -2x^2 + 3x + 5

h is the height above the ground the shot reaches as a function of the horizontal distance it travels.

Solve -2x^2 + 3x + 5 = 0 for x.

-2x^2 + 3x + 5 = 0

2x^2 - 3x - 5 = 0

(2x - 5)(x + 1) = 0

2x - 5 = 0 or x + 1 = 0

2x = 5 or x = -1

x = 2.5 or x = -1

At x = 2.5 ft, the height was 0. That is when the shot fell on the ground.

From the zero of the function at x = 2.5 ft, we find out the shot traveled 2.5 ft horizontally.

Answer: A. the horizontal distance covered by the shot

Answer 2

Hi!

So the original formula for a problem likes this or a quadratic is:

ax^2 + bx+ c

so a is the accelaration; b is the initial velocity; and c is the initial height

so none of those are the zeros, which means c is out

zeros are the x values when y is 0, that also means when the parabola crosses the x axis

so d is out too because the maximum height is the vertex which mean that there would probably be a y value. in some cases the vertex is the zero, but not here

B is out because x isn't time it is distance

your answer is A!

Hope this helps!