***I USED A DIFFERENT PROBLEM. FOLLOW THE STEPS!!!
The vertex is either the minimum point of the graph of the maximum point of the graph if the equation is a quadratic equation.
if it is the minimum point on the graph, than any other point on the graph will be above it.
if it is the maximum point on the graph, than any other point on the graph will be below it.
(-3,125) is above (0,8).
(0,8) is therefore a minimum point on the graph and the graph opens up.
in fact, your equation is y = 13x^2 + 8.
when x = 0, y = 8 as you can see by replacing x with 0 and getting 13*0^2 + 8 = 8 which becomes 8 = 8.
when x = -3, your equation becomes y = 13 * (-3)^2 + 8 which becomes y = 13 * 9 + 8 which becomes y = 117 + 8 which becomes y = 125.
when x = -3, y = 125 which corresponds to the REQUIREMENTS that the point (-3,125) be on the grpah of the equation.
your equation is shown below:
graph%28600%2C600%2C-4%2C4%2C-5%2C200%2C13x%5E2+%2B+8%2C8%2C125%29
i drew horizontal lines at y = 8 and y = 125.
the intersection of those lines with the graph gives you the value of x when you draw a vertical line from the intersection to the x-axis.
Please help, I am really having trouble understanding this problem.
what are the foci of the hyperbola given by the equation 16y^2-9x^2=144
1 answer
1 answer