The path of a football kicked by a field goal kicker can be modeled by the equation y = –0.03x2 + 1.53x, where x is the horizontal distance in yards and y is the corresponding height in yards.

What is the football’s maximum height? Round to the nearest tenth.
yds.
How far is the football kicked?
yds.

it has the same slope and a different y-intercept

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24,18t

step-by-step explanation:

part a: triangles gcb and epb

part b. sides cg and rp are parallel by definition of parallelogram.

angles cgb and e are congruent by being alternate interior angles of parallel lines cut by a transversal. angles gbc and ebp are congruent because they are vertical angles. by aa similarity, the triangles are similar.

part c. since the triangles are similar, the ratios of the lengths of corresponding sides are equal.

calculation of be:

gb/be = cb/bp

(400 ft)/be = 300/200

(400 ft)/be = 3/2

3be = 2 * 400 ft

3be = 800 ft

be = 800/3 ft

calculation of pe:

gc/pe = cb/bp

(350 ft)/pe = 300/200

(350 ft)/pe = 3/2

3pe = 2 * 350 ft

3pe = 700 ft

pe = 700/3 ft

4.28095238095 that is the answer