Question

a blue jay swoops down from the top of a 10m tree to chase away some house sparrows. the blue jay’s path follows a parabolic path given by the function $h(t) = 2t^2 - 8t + 10$ where $t$ is time in seconds and $h(t)$ is height in metres.

b. graphically determine the times when the blue jay reaches a height of 4m.

Explanation:

Step1: Analyze the function

The function is \( h(t) = 2t^2 - 8t + 10 \), a quadratic (parabolic) function. To graph it, we can find vertex, y - intercept, and maybe a few other points. The vertex of a parabola \( ax^2+bx + c \) is at \( t=-\frac{b}{2a} \). Here, \( a = 2 \), \( b=-8 \), so \( t =-\frac{-8}{2\times2}= 2 \). Substitute \( t = 2 \) into \( h(t) \): \( h(2)=2\times(2)^2-8\times(2)+10=8 - 16 + 10 = 2 \). So the vertex is at \( (2,2) \). The y - intercept (when \( t = 0 \)) is \( h(0)=10 \), so the point is \( (0,10) \).

Step2: Graph the line \( h(t)=4 \)

We also need to graph the horizontal line \( y = 4 \) (since we want to find when \( h(t)=4 \)).

Step3: Find intersection points

The parabola \( h(t)=2t^2 - 8t + 10 \) and the line \( y = 4 \) will intersect at two points (since the parabola opens upwards as \( a = 2>0 \) and the vertex is at \( y = 2<4 \)). To find the times, we can also solve the equation \( 2t^2-8t + 10=4 \) algebraically (to check) : \( 2t^2-8t + 6 = 0 \), divide by 2: \( t^2-4t + 3=0 \), factor: \( (t - 1)(t - 3)=0 \), so \( t = 1 \) or \( t = 3 \). Graphically, we plot the parabola with vertex (2,2), passing through (0,10), and the line \( y = 4 \). The points where they intersect will have t - values of 1 and 3.

Answer:

The blue jay reaches a height of 4m at \( t = 1 \) second and \( t = 3 \) seconds.