What is the vertex and the x-intercepts of the equation y=300x-16x²?

Answer 1

y = 300x - 16x²
The general form of a quadratic equation is ax² + bx +c = 0.
Because there is no number value (c), this implies the x-intercept is zero.
The vertex point, at that point the slope of the graph will be zero. So use calculus, put the equation equal to zero and get x and y values to locate your point.
y = 300x - 16x²
dy/ dx = 300 - 32x
300 - 32x = 0
32x = 300
x = 300/ 32
x = 9∙375

y = 300x - 16x²
y = 300(9∙375) - 16(9∙375)²
y = 300(9∙375) - 16(9∙375)²
y = 2812∙5 - 1406∙25
y = 1406∙25

Vertex pt. (9∙375, 1406∙25).

Answer 2

f `(x) = 300 - 32 x = 0 for turning point
32 x = 300
x = 300 / 32
x = 75 / 8
y = 300 (75/8) - 16 (75/8)²
y = (75/8) (300 - 150)
y = (75/8) (150)
y = 75² / 4
y = 5625 / 4
Vertex (75/8 , 5625/4)
For x intercepts:-
300x - 16x² = 0
16 x ² - 300 x = 0
4 x ² - 75 x = 0
x (4x - 75) = 0
x = 0 , x = 75 / 4

Answer 3

here how to find the vertex and the x intercept using your calculator