ten in pen, except for graphs and drawings, $\newline$ While playing golf Laura hit her ball from the ground. The height, in feet, of her golf ball can be modeled by $h(t)=-16 t^{2}+48 t$ , where $t$ is the time in seconds. $\newline$ Graph $h(t)$ on the set of axes below.

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Solve word problems by finding two-variable equations

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Q. ten in pen, except for graphs and drawings,

\newline

While playing golf Laura hit her ball from the ground. The height, in feet, of her golf ball can be modeled by

h(t)=-16 t^{2}+48 t

, where

t

is the time in seconds.

\newline

Graph

h(t)

on the set of axes below.

Find y-intercept: To graph the function $h(t) = -16t^2 + 48t$ , we need to determine the key features of the parabola, such as the vertex, the axis of symmetry, and the y-intercept.
Calculate vertex: First, let's find the y-intercept by evaluating $h(t)$ when $t = 0$ . $\newline$ $h(0) = -16(0)^2 + 48(0) = 0$ $\newline$ The y-intercept is at the point $(0, 0)$ .
Determine axis of symmetry: Next, we find the vertex of the parabola. The vertex form of a parabola is $h(t) = a(t-h)^2 + k$ , where $(h, k)$ is the vertex. For the given function, we can find the time $t$ at which the vertex occurs by using the formula $t = -\frac{b}{2a}$ , where $a = -16$ and $b = 48$ . $\newline$ $t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5$
Sketch graph: Now we calculate the height $h$ at the time $t = 1.5$ seconds to find the vertex. $\newline$ $h(1.5) = -16(1.5)^2 + 48(1.5) = -16(2.25) + 72 = -36 + 72 = 36$ $\newline$ The vertex is at the point $(1.5, 36)$ .
Calculate additional points: The axis of symmetry is a vertical line that passes through the vertex. Since the vertex is at $t = 1.5$ , the axis of symmetry is the line $t = 1.5$ .
Calculate additional points: The axis of symmetry is a vertical line that passes through the vertex. Since the vertex is at $t = 1.5$ , the axis of symmetry is the line $t = 1.5$ .Now we can sketch the graph. We start by plotting the y-intercept $(0, 0)$ and the vertex $(1.5, 36)$ . We know that the parabola opens downward because the coefficient of $t^2$ is negative ( $-16$ ). We also plot the axis of symmetry at $t = 1.5$ .
Calculate additional points: The axis of symmetry is a vertical line that passes through the vertex. Since the vertex is at $t = 1.5$ , the axis of symmetry is the line $t = 1.5$ .Now we can sketch the graph. We start by plotting the y-intercept $(0, 0)$ and the vertex $(1.5, 36)$ . We know that the parabola opens downward because the coefficient of $t^2$ is negative $(-16)$ . We also plot the axis of symmetry at $t = 1.5$ .To get more points for the graph, we can choose values of $t$ to the left and right of the vertex and calculate the corresponding $h(t)$ values. For example, let's calculate $h(1)$ and $t = 1.5$ $0$ . $h(1) = -16(1)^2 + 48(1) = -16 + 48 = 32$ $h(2) = -16(2)^2 + 48(2) = -16(4) + 96 = -64 + 96 = 32$ We have two more points: $t = 1.5$ $1$ and $t = 1.5$ $2$ .
Calculate additional points: The axis of symmetry is a vertical line that passes through the vertex. Since the vertex is at $t = 1.5$ , the axis of symmetry is the line $t = 1.5$ .Now we can sketch the graph. We start by plotting the y-intercept $(0, 0)$ and the vertex $(1.5, 36)$ . We know that the parabola opens downward because the coefficient of $t^2$ is negative $(-16)$ . We also plot the axis of symmetry at $t = 1.5$ .To get more points for the graph, we can choose values of $t$ to the left and right of the vertex and calculate the corresponding $h(t)$ values. For example, let's calculate $h(1)$ and $t = 1.5$ $0$ . $h(1) = -16(1)^2 + 48(1) = -16 + 48 = 32$ $h(2) = -16(2)^2 + 48(2) = -16(4) + 96 = -64 + 96 = 32$ We have two more points: $t = 1.5$ $1$ and $t = 1.5$ $2$ .Using the points $(0, 0)$ , $t = 1.5$ $1$ , $(1.5, 36)$ , and $t = 1.5$ $2$ , we can draw a symmetrical parabola on the graph, ensuring that it opens downward and passes through these points.