4. Gloria the grasshopper is working on her hops.
She is trying to jump as high and as far as she
can. Her best jump so far was 28 cm long,
and she reached a height of 20 cm. Sketch a
graph and write an equation of the parabola
that describes the path of her jump.(5pts)


Sagot :

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola  that describes the path of her jump is [tex]\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}[/tex]

The given parameters are:

[tex]\mathbf{Height = 20}[/tex]

[tex]\mathbf{Length = 28}[/tex]

Assume she starts from the origin (0,0)

The midpoint would be:

[tex]\mathbf{Mid = \frac 12 \times Length}[/tex]

[tex]\mathbf{Mid = \frac 12 \times 28}[/tex]

[tex]\mathbf{Mid = 14}[/tex]

So, the vertex of the parabola is:

[tex]\mathbf{Vertex = (Mid,Height)}[/tex]

Express properly as:

[tex]\mathbf{(h,k) = (14,20)}[/tex]

A point on the graph would be:

[tex]\mathbf{(x,y) = (28,0)}[/tex]

The equation of a parabola is calculated using:

[tex]\mathbf{y = a(x - h)^2 + k}[/tex]

Substitute [tex]\mathbf{(h,k) = (14,20)}[/tex] in [tex]\mathbf{y = a(x - h)^2 + k}[/tex]

[tex]\mathbf{y = a(x - 14)^2 + 20}[/tex]

Substitute [tex]\mathbf{(x,y) = (28,0)}[/tex] in [tex]\mathbf{y = a(x - 14)^2 + 20}[/tex]

[tex]\mathbf{0 = a(28 - 14)^2 + 20}[/tex]

[tex]\mathbf{0 = a(14)^2 + 20}[/tex]

Collect like terms

[tex]\mathbf{a(14)^2 =- 20}[/tex]

Solve for a

[tex]\mathbf{a =- \frac{20}{14^2}}[/tex]

[tex]\mathbf{a =- \frac{20}{196}}[/tex]

Simplify

[tex]\mathbf{a =- \frac{5}{49}}[/tex]

Substitute [tex]\mathbf{a =- \frac{5}{49}}[/tex] in [tex]\mathbf{y = a(x - 14)^2 + 20}[/tex]

[tex]\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}[/tex]

Hence, the equation of the parabola  that describes the path of her jump is [tex]\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}[/tex]

See attachment for the graph

Read more about equations of parabola at:

https://brainly.com/question/4074088

View image MrRoyal