To find the vertex of the quadratic function given by the equation:
\[ h(t) = -6t^2 - 24t + 20 \]
We can use the formula for the vertex of a quadratic function in standard form \(h(t) = at^2 + bt + c\), where the vertex \(t\)-coordinate is given by:
\[ t = -\frac{b}{2a} \]
In this case, the coefficients are:
- \(a = -6\)
- \(b = -24\)
Now, let's calculate the \(t\)-coordinate of the vertex:
\[ t = -\frac{-24}{2 \cdot -6} = \frac{24}{-12} = -2 \]
Next, we need to find the \(h\)-coordinate of the vertex by substituting \(t = -2\) back into the function:
\[ h(-2) = -6(-2)^2 - 24(-2) + 20 \] \[ h(-2) = -6(4) + 48 + 20 \] \[ h(-2) = -24 + 48 + 20 \] \[ h(-2) = 24 \]
So the vertex is at the point \((-2, 24)\).
Now, let's look at the options:
A) (44, −2)
B) (1, 0)
C) (−2, 44)
D) (−5, 0)
None of the given options appear to fit the vertex we found. However, if we clarify that the ordered pair of interest should indeed represent the vertex, then none of the answer options are correct.
The correct ordered pair for the vertex of the function \(h(t) = -6t^2 - 24t + 20\) is \((-2, 24)\).