Question

3.(08.07 HC)

Answer 1

Answer:

For a general equation:

y = f(x) =  a*x^2 + b*x + c

The x-intercepts are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4*a*c} }{2*a}$

And the vertex is the point (h, k) such that:

h = -b/2*a

k = f(h)

Part A)

We have the function f(x) = -16*x^2 + 22*x + 3

The x-intercepts are then:

$x = \frac{-(22) \pm \sqrt{22^2 - 4*(-16)*3} }{2*(-16)} = \frac{-22 \pm 26 }{-32}$

The two intercepts are:

x = (-22 - 26)/(-32) = 1.5

x = (-22 + 26)/(-32) = -0.125

Part B:

The vertex is (h, k), such that:

h = -22/(2*(-16)) = -22/-32 = 0.6875

k = f(0.6875) = -16*(0.6875)^2 + 22*0.6875 + 3 = 10.6525

Then the vertex is: ( 0.6875, 10.6525)

If the leading coefficient is positive, then the vertex is a minimum

If the leading coefficient is negative, then the vertex is a maximum.

In this case the leading coefficient is -16, then we can conclude that the vertex is a maximum.

Part C:

To graph the function, we can graph the points that we already know (the vertex and the two x-intercepts) and connect them with a curve.

You could also add another few points so you have a guide to draw the curve, for example the point:

y = f(0) = -16*0^2 + 22*0 + 3

    f(0) = 3

(0, 3)