I need help on 4 quadratics and turn them into graghs with a form of a parbola

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1 Answer

We find the x-intercepts by setting y to zero.

-x^2 + 35x = 0
x^2 - 35x = 0

We use "completing the square."

x^2 - 35x + (35/2)^2 = (35/2)^2
x^2 - 35x + 17.5^2 = 17.5^2
(x - 17.5)^2 = 17.5^2
x - 17.5 = ±17.5
x = 17.5 ±17.5
x = 17.5 + 17.5    and    x = 17.5 - 17.5
x = 35   and    x = 0

The x-intercepts are (0, 0)  and (35, 0)
The vertex has an x value halfway between them: 17.5
We can calculate the y value of the vertex using x=17.5

y = -x^2 + 35x
y = -(17.5)^2 + 35(17.5)
y = -306.25 + 612.5
y = 306.25

The vertex is at (17.5, 306.25)

Graphing this equation would show a parabola opening downward,
with the highest point at (17.5, 306.25), crossing the x-axis
at (0, 0) and (35, 0).

 

Problem statement: y=-x^2+35x
I need help on 4 quadratics and turn them into graghs with a form of a parbola
by Level 11 User (78.4k points)

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