Maths-

General

Easy

Question

# Suppose the path of the ball in above figure is f(x) = -0.25(x-1)^{2} + 6.25. Find the ball’s initial and maximum heights.

Hint:

### For a quadratic function is in standard form, f(x)=ax2+bx+c.

The standard quadratic form is ax2+bx+c=y, In this the term c gives us the y-intercept of the curve.This is the point at which ball’s initial height.

The vertex form of a quadratic equation is y=a(x−h)2+k. This gives the vertex (h, k) in which y coordinate is the top most point of the ball.

## The correct answer is: C = 6

### The given equation is f(x) = -0.25(x-1)^{2} + 6.25

This is in the vertex form

Therefore on comparing with the vertex form of a quadratic equation is y=a(x−h)^{2}+k. we get, h = 1 , k = 6.25

Therefore, vertex is (h,k) = (1, 6.25)

So the maximum height of the ball is 6.25 .

Converting the equation in standard form

f(x) = -0.25(x-1)^{2} + 6.25

f(x) = -0.25(x^{2}+1 – 2x) + 6.25

f(x) = -0.25x^{2} – 0.25 + 0.5x + 6.25

f(x) = -0.25x^{2} +0.5 x + 6.00

Therefore, y intercept is c = 6

Which is the initial height of the ball i.e 6

^{2}+ 6.25

^{2}+1 – 2x) + 6.25

^{2}– 0.25 + 0.5x + 6.25

^{2}+0.5 x + 6.00

Therefore, y intercept is c = 6

Which is the initial height of the ball i.e 6

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### Mia tosses a a ball to her dog. The function -0.5(x-2)^{2} + 8 represents the ball path.

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