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# The function h$t$=-5t^2+20t+60 models the height in meters of a ball t seconds after it’s thrown . Which describe the intercepts and vertex of this function

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## Answer to a math question The function h$t$=-5t^2+20t+60 models the height in meters of a ball t seconds after it’s thrown . Which describe the intercepts and vertex of this function

Santino
4.5
To find the intercepts of the function h$t$ = -5t^2 + 20t + 60, we need to set it equal to zero and solve for t.

1. Set h$t$ = 0:
-5t^2 + 20t + 60 = 0

2. Solve the quadratic equation using the quadratic formula:
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a = -5, b = 20, and c = 60. Plugging these values into the quadratic formula:

t = \frac{-20 \pm \sqrt{20^2 - 4$-5$$60$}}{2$-5$}

3. Simplify the equation:
t = \frac{-20 \pm \sqrt{400 + 1200}}{-10}
t = \frac{-20 \pm \sqrt{1600}}{-10}
t = \frac{-20 \pm 40}{-10}

4. Solve for t:
Case 1:
t = \frac{-20 + 40}{-10}
t = -2

Case 2:
t = \frac{-20 - 40}{-10}
t = 6

Therefore, the intercepts of the function h$t$ are t = -2 and t = 6. This means that the graph of the function crosses the t-axis at t = -2 and t = 6.

To find the vertex of the function, we can use the formula for the x-coordinate of the vertex, which is given by:

x = -\frac{b}{2a}

In this case, a = -5 and b = 20. Plugging these values into the formula:

x = -\frac{20}{2$-5$}
x = -\frac{20}{-10}
x = 2

To find the y-coordinate of the vertex, we substitute the x-coordinate back into the function:

h$2$ = -5$2$^2 + 20$2$ + 60
h$2$ = -20 + 40 + 60
h$2$ = 80

Therefore, the vertex of the function h$t$ is $2, 80$.

The intercepts of the function h$t$ = -5t^2 + 20t + 60 are t = -2 and t = 6.
The vertex of the function h$t$ is $2, 80$.
Frequently asked questions $FAQs$
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