Question 6.10.7: Domain and Range of a Quadratic Function a) Sketch the graph......

a) Since f (x) means y, we can replace f (x) with y to obtain y = -2x² + 3x + 4.

Now graph y = -2x² + 3x + 4 using the steps outlined in the general procedure.

1. Since a = -2, which is less than 0, the parabola opens downward.

2. Axis of symmetry:

x = \frac{-b}{2a} = \frac{-(3)}{2(-2)} = \frac{-3}{-4} = \frac{3}{4}

Thus, the axis of symmetry is x = \frac{3}{4}.

3. y-coordinate of vertex:

y = -2x² + 3x + 4

= -2\left ( \begin{matrix} \frac{3}{4} \end{matrix} \right )² + 3\left ( \begin{matrix} \frac{3}{4} \end{matrix} \right ) + 4

= -2\left ( \begin{matrix} \frac{9}{16} \end{matrix} \right ) + \frac{9}{4} + 4

= -\frac{9}{8} + \frac{9}{4} + 4

= -\frac{9}{8} + \frac{18}{8} + \frac{32}{8} = \frac{41}{8}     or     5\frac{1}{8}

Thus, the vertex is at (\frac{3}{4}, 5\frac{1}{8}).

4. y-intercept: y = -2x² + 3x + 4

= -2(0)² + 3(0) + 4 = 4

Thus, the y-intercept is (0, 4).

5. x-intercepts: y = -2x² + 3x + 4

0 = -2x² + 3x + 4     or     -2x² + 3x + 4 = 0

This equation cannot be factored, so we will use the quadratic formula to solve it.

a = -2,     b = 3,     c = 4

x = \frac{-b  ±  \sqrt{b²  –  4ac}}{2a}

= \frac{-3  ±  \sqrt{3²  –  4(-2)(4)}}{2(-2)}

= \frac{-3  ±  \sqrt{9  +  32}}{-4}

= \frac{-3  ±  \sqrt{41}}{-4}

Since\sqrt{41} ≈ 6.4,

x ≈ \frac{-3  +  6.4}{-4} ≈ \frac{3.4}{-4} ≈ -0.85     or     x ≈ \frac{-3  –  6.4}{-4} ≈ \frac{-9.4}{-4} ≈ 2.35

Thus, the x-intercepts are about (-0.85, 0) and (2.35, 0).

6. Plot the vertex (\frac{3}{4}, 5\frac{1}{8}) , the y-intercept (0, 4), and the x-intercepts (-0.85, 0) and (2.35, 0). Then sketch the graph (Fig. 6.47).

b) The domain, the values that can be used for x, is the set of all real numbers, \mathbb{R}.

The range, the values of y, is y ≤ 5\frac{1}{8}.