a soccer ball is kicked from the ground. After travelling 10m, it just clears a

5 `11/14` meter tall fence. it strikes the ground 37 meters from the start point. the path is shown below.

express the height of the ball as a function of the distance from the kicking point in factored, vertex and standard form. All constants should be rounded to two decimals

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We have three points: (0,0),(10,`81/14` ),and (37,0).

The three forms of the quadratic equation we are asked to find are:

**factored form** y=a(x-p)(x-q) where a is a vertical stretch factor, and p,q are the zeros.

**standard form** `y=ax^2+bx+c`

**vertex form** `y=a(x-h)^2+k` where a is the vertical stretch factor, and (h,k) is the vertex.

Here y is the vertical height and x the horizontal distance.

(a) Factored form -- we have the two zeros so p=0 and q=37. Then y=a(x-0)(x-37). We can find a using the third point given -- substitute 10 for x and `81/14` for y and we get:

`81/14=a(10)(10-37)`

`81/14=-270a`

`a=(81/14)/(-270)=-3/140`

So the equation we seek is `y=-3/140x(x-37)`

(b) We can expand the product to put in standard form:

`y=-3/140x^2+111/140x`

(c) We can complete the square to put in vertex form:

Factor out the leading coefficient:

`y=-3/140(x^2-37x)` Add and subtract `(37/2)^2=1369/4`

`y=-3/140(x^2-37x+1369/4-1369/4)`

`y=-3/140((x-37/2)^2-1369/4)`

`y=-3/140(x-37/2)^2+4107/560`

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With y the height of the ball as a function of x, the horizontal distance, we have:

factored form: `y=-.02x(x-37)`

standard form: `y=-.02x^2+.79x`

vertex form: `y=-.02(x-18.5)^2+7.33`

All units in meters and fractions rounded to 2 decimal places.

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The graph:

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