Question

How do you graph the parabola `h(t)=-16t^2+280t+17` using vertex, intercepts and additional points?

Answer 1

Graph parabola

Explanation:

The main features to graph this parabola are:
1. It is downward because a < 0
2. Vertex.
x-coordinate of vertex:
`x = -b/(2a) = -280/-32 = 35/4`
y-coordinate of vertex:
`y(35/4) = - 16(1225/16) + 280(35/4) + 17`
3. y-intercept.
Make x = o, y-intercept --> 17
4. x-intercepts. Make y = 0 andsSolve the quadratic function by the new improved quadratic formula in graphic form (Socratic Search)
`y = -16x^2 + 280x + 17 = 0`
`D = d^2 = b^2 - 4ac = 32,400 + 1088 = 33,488` --> `d = +- 183`
There are 2 x-intercepts (real roots);
`x = -b/(2a) +- d/(2a) = -280/32 +- 183/32 = 35/4 +- 183/32`