This question is incomplete, the complete question is;
A javelin is thrown in the air. Its height is given by h(x) = -1/20x² + 8x + 6
where x is the horizontal distance in feet from the point at which the javelin is thrown.
a. How high is the javelin when it was thrown?
b. What is the maximum height of the javelin?
c. How far from the thrower does the javelin strike the ground?'
Answer:
a. height of the javelin when it was thrown is 6 ft
b. the maximum height of the javelin is 326 ft
c. distance from the thrower is 160.75 ft
Explanation:
a)
Given h(x) = -1/20x² + 8x + 6
we determine the height when x = 0
h(0) = -1/20(0)² + 8(0) + 6 = 6 ft
therefore height of the javelin when it was thrown is 6 ft
b)
to determine the maximum height of the javelin;
we find the vertex of the quadratic
so
h = - [ 8 / ( 2(-1/20) ) ] = 80
therefore
h(80) = -1/20(80)² + 8(80) + 6
= -320 + 640 + 6 = 326 ft
therefore the maximum height of the javelin is 326 ft
c)
Now the thrower is at the point ( 0,0 ) and the javelin comes down at another point ( x,0 )
this is possible by calculating h(x) = 0
⇒ -1/20x² + 8x + 6 = 0
⇒ x² - 160x - 120 = 0
⇒ x = [ -(-160) ± √( (-160)² - 4(1)(-120) ) ] / [ 2(1) ]
x = [ 160 ± √(25600 + 480) ] / 2
so
[x = 160.75 ; x = -0.75 ]
distance cannot be Negative
therefore distance from the thrower is 160.75 ft
The maximum height, the location on the ground and the initial vertical height of the javelin is required.
The initial height of the javelin is 6 feet.
The maximum height of the javelin is 326 feet.
The javelin strikes the ground at 160.75 feet.
The given equation is
[tex]h(x)=-\dfrac{1}{20}x^2+8x+6[/tex]
where [tex]x[/tex] is the horizontal distance
At [tex]x=[/tex] we will get the initial vertical height.
[tex]h(0)=-\dfrac{1}{20}\times0+8\times 0+6\\\Rightarrow h(0)=6[/tex]
Vertex of a parabola is given by
[tex]x=-\dfrac{b}{2a}=-\dfrac{8}{2\times -\dfrac{1}{20}}\\\Rightarrow x=80[/tex]
[tex]h(80)=-\dfrac{1}{20}(80)^2+8\times 80+6=326[/tex]
At [tex]h(x)=0[/tex] the javelin will hit the ground
[tex]0=-\dfrac{1}{20}x^2+8x+6\\\Rightarrow -x^2+160x+120=0\\\Rightarrow x=\dfrac{-160\pm \sqrt{160^2-4\left(-1\right)\times120}}{2\left(-1\right)}\\\Rightarrow x=-0.75,160.75[/tex]
Learn more about parabolas from:
https://brainly.com/question/14477557
