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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.

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This is the equation of an upside down parabola, so the maximum height will be at the vertex. To find the vertex, we'll use the formula for the x-coordinate and then solve for the y-coordinate.
[tex]x=\frac{-1.53}{2*-.03}=25.5[/tex]

Plugging this into the given formula:
[tex]y=-0.03(25.5)^2+1.53(25.5)=19.5075[/tex]
Since it asks us to round, the maximum height would be 20 yards. To find how far the football travels in total, we need to set the equation equal to 0 and find the roots:
[tex]0=-0.03x^2+1.53x=x(-0.03x+1.53)[/tex]

Then by the zero product property (or the no zero divisors rule), we have:
[tex]x=0[/tex] (this is the x value where the ball is kicked from)
[tex]-0.03x+1.53=0[/tex]
[tex]-0.03x=-1.53[/tex]
[tex]x=51[/tex]

So the ball travels 51 yards in total.

Answer:

Maximum height achieved by football = 25.5 yards.

Distance traveled by football  51 yards.

Step-by-step explanation:

Path of a football kicked by a field goal kicker can be modeled by the equation y = -0.03x² + 1.53x.

Where x represents horizontal distance and y represents height of the football.

For maximum, height we will find the [tex]\frac{dy}{dx}[/tex] ( derivative ) of the equation and equate it to zero.

[tex]\frac{d(y)}{dx}[/tex] = [tex]\frac{d}{dx}[/tex] (-0.03x² + 1.53x )

[tex]\frac{dy}{dx}[/tex] = -0.03 × 2x + 1.53

By putting [tex]\frac{dy}{dx}[/tex] = 0

-0.06x + 1.53 = 0

0.06x = 1.53

x = [tex]\frac{1.53}{0.06}[/tex] = 25.5 yards

Therefore, maximum height achieved by the ball is  25.5 yards.

Now for horizontal distance covered by ball y should be 0

-0.03x² + 1.53x = 0

x ( 1.53 - 0.03x ) = 0

0.03x = 1.53

x = [tex]\frac{1.53}{0.03}[/tex] = 51 yards.

Football was kicked 51 yards.

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