Quadratic formula ?

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2 Answers
Mar 14, 2018

Alex will land #3.9# meters from the ramp.

Explanation:

We seek to find #h(d) = 0#.

#0 = -3.9d^2 + 13.1d + 8.7#

Now apply the quadratic formula.

#d = (-13.1 +- sqrt(13.1^2 - 4 * -3.9 * 8.7))/(2 * -3.9)#

#d = (-13.1 +- sqrt(307.33))/(-7.8)#

We now use a calculator.

#d = 3.927 m or -0.568# m

A negative answer is clearly impossible so Alex will land #3.9# meters away from the ramp. The graph of the #h(d)# function confirms.

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Hopefully this helps!

Mar 14, 2018

#19.7# metres in height.

Explanation:

It's stated in this problem that Alex's path can be modelled by the quadratic function #h(d) = - 3.9d^(2) + 13.1d + 8.7#.

The maximum value of this function will be the maximum height reached by Alex.

In order to find the maximum value of #h(d)#, we need to complete the square:

#Rightarrow h(d) = - 3.9d^(2) + 13.1d + 8.7#

#Rightarrow h(d) = - 3.9 (d^(2) + frac(13.1)(- 3.9)d + frac(8.7)(- 3.9))#

#Rightarrow h(d) = - 3.9 (d^(2) - frac(13.1)(3.9)d - frac(8.7)(3.9))#

#Rightarrow h(d) = - 3.9 (d^(2) - frac(13.1)(3.9)d + (frac(frac(13.1)(3.9))(2))^(2) - frac(8.7)(3.9) - (frac(frac(13.1)(3.9))(2))^(2))#

#Rightarrow h(d) = - 3.9 (d^(2) - frac(13.1)(3.9)d + (frac(13.1)(7.8))^(2) - frac(8.7)(3.9) - (frac(13.1)(7.8))^(2))#

#Rightarrow h(d) = - 3.9 ((d- frac(171.61)(60.84))^(2) -5.0514464169)#

#Rightarrow h(d) = - 3.9 (d- frac(171.61)(60.84))^(2) + 19.70064103#

Now, the expression #(d- frac(171.61)(60.84))^(2)# is always positive, as it is squared, i..e it is always greater than or equal to zero.

#Rightarrow (d- frac(171.61)(60.84))^(2) geq 0#

Multiplying this expression by #- 3.9# (a negative number) reverses the inequality:

#Rightarrow - 3.9 (d- frac(171.61)(60.84))^(2) leq 0#

Let's add #19.70064103# to both sides of the inequality:

#Rightarrow - 3.9 (d- frac(171.61)(60.84))^(2) leq 19.70064103#

This expression is now equivalent to our function #h(d)# (in vertex form):

#therefore h(d) leq 19.70064103#

According to this inequality, the function #h(d)# is less than or equal to #19.70064103#, i.e. has a maximum value of #19.70064103#.

Rounding this number to the nearest tenth gives #h(d) leq 19.7#.

Therefore, Alex reaches no higher than #19.7# metres.