Math

QuestionAn athlete releases a shot put at 6565^{\circ}. Its height is modeled by f(x)=0.03x2+2.1x+6.1f(x)=-0.03 x^{2}+2.1 x+6.1. Find the max height and distance from release.

Studdy Solution

STEP 1

Assumptions1. The height of the shot, f(x)f(x), is given by the equation f(x)=0.03x+.1x+6.1f(x)=-0.03x^{}+.1x+6.1 . xx represents the shot's horizontal distance, in feet, from its point of release

STEP 2

The maximum height of the shot occurs at the vertex of the parabola represented by the equation f(x)=0.03x2+2.1x+6.1f(x)=-0.03x^{2}+2.1x+6.1. The x-coordinate of the vertex of a parabola given in the form f(x)=ax2+bx+cf(x)=ax^{2}+bx+c is given by the formula b2a-\frac{b}{2a}.

STEP 3

Substitute the values of aa and bb from the equation f(x)=0.03x2+2.1x+6.1f(x)=-0.03x^{2}+2.1x+6.1 into the formula b2a-\frac{b}{2a} to find the x-coordinate of the vertex.
xvertex=b2a=2.12×0.03x_{vertex} = -\frac{b}{2a} = -\frac{2.1}{2 \times -0.03}

STEP 4

Calculate the x-coordinate of the vertex.
xvertex=2.12×0.03=35x_{vertex} = -\frac{2.1}{2 \times -0.03} =35

STEP 5

The x-coordinate of the vertex represents the horizontal distance from the point of release at which the maximum height occurs. Therefore, the maximum height occurs35 feet from the point of release.

STEP 6

To find the maximum height of the shot, substitute the x-coordinate of the vertex into the equation f(x)=0.03x2+2.1x+6.1f(x)=-0.03x^{2}+2.1x+6.1.
f(35)=0.03(35)2+2.1(35)+6.1f(35) = -0.03(35)^{2} +2.1(35) +6.1

STEP 7

Calculate the maximum height of the shot.
f(35)=0.03(35)2+2.1(35)+6.1=40.85f(35) = -0.03(35)^{2} +2.1(35) +6.1 =40.85The maximum height of the shot is40.85 feet, which occurs35 feet from the point of release.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord