(1) Attached figure 1 is the solution
By graphing <span>the model of the data ⇒⇒⇒ red points
</span><span>(-1,0), (1,-4), (2,-3), (4,5), (5,12)
</span><span>and graphing the quadratic equations
∴ the solution will be choice (d)</span>
<span>d.] f(x) = (x-1)² - 4
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(2) </span><span>Attached figure 2 is the solution
</span><span>By graphing <span>the model of the data ⇒⇒⇒ red points
</span></span><span>(0,1), (1,4), (2,5), (3,4), (4,1)
</span>and graphing the quadratic equations
∴ the solution will be choice (c)
<span>c.] f(x) = – (x-2)² + 5
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</span>
<span>problems (1),(2) can be solved by substituting with the model of data
at the general form the quadratic function:
F(x) = a (x+b)² + c
and solve to find a, b, c
OR, It is easier to use the graph to find the <span>appropriate quadratic equation
as we did previously
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</span>(3) </span>
h(t) = - 12t² + 36t.
<span> when the object hits the ground ⇒⇒ h = 0
∴ </span><span>- 12t² + 36t = 0 ⇒⇒ factor t
∴ t ( -12t +36 ) = 0
t = 0 (unacceptable) OR t = 36/12 = 3 sec.
</span><span>∴ the solution will be choice (c)
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(4)
</span><span>h(t) = - 8t² + 32t
At maximum height ⇒⇒⇒ dh/dt = 0
∴ dh/dt = -16t +32 = 0
∴ t = 32/16 = 2 sec.
</span>
∴ the solution will be choice (b)