y = 4.5x² + 9x + 15
4.5 x^2 + 9 x = y - 15
x^2 + 2 x = (10/45)(y-15)
x^2 + 2 x +(2/2)^2 = (10/45)(y-15) + 1
You have a sign error, the height goes to + infinity for large + or - x
I will lay it out your way but suspect
h = initial height = 15 m
Vi = initial speed up = 9 m/s
4.5 = 9.81/2 = g/2 = half gravity
so
h = Hi + Vi t - (g/2)t^2
is here
h = 15 + 9 t - 4.9 t^2
h = 0 is hit the water
One of your times will be negative, that was before you jumped :)
but anyway using what you have:
(x+1)^2 = (10/45)( y-15 ) + (10/45)(45/10)
(x+1)^2 = (10/45 ( y -15+45/10) = (10/45)(y -10.5)
vertex at (- 1, 10.5) symmetric about x = -1
for x axis crossings put y = 0 in
x+1 = /-(10/45) sqrt (-10.5), imaginary, always above axis of course
when x = 0, y axis intercept = 15
now do it right with gravity down :)
47. The equation y = 4.5x² + 9x + 15 describes the height of a diver, in metres, at x seconds.
Change this equation into vertex form by completing the square. State the y-intercept, the equation of the axis of symmetry, and the vertex.
B. Before locating the x-intercepts of the parabola, state the number of roots of the equation.
C. Use the quadratic equation formula to determine the x-intercepts of the parabola when the diver hits the water.
D. Explain why one root is inadmissible. Make a sketch to represent the path of the diver.
1 answer
1 answer