Answer:
[tex]\boxed{\text{B. x = 4 and x = -1.75}}[/tex]
Step-by-step explanation:
ƒ(x) = x² - 2x – 8; g(x) = ¼x -1
If ƒ(x) = g(x), then
x² - 2x – 8 = ¼x -1
One way to solve this problem is by completing the square.
Step 1. Subtract ¼ x from each side
[tex]x^{2} - \dfrac{9}{4}x - 8 = -1[/tex]
Step 2. Move the constant term to the other side of the equation
[tex]x^{2} - \dfrac{9}{4}x = 7[/tex]
Step 3. Complete the square on the left-hand side
Take half the coefficient of x, square it, and add it to each side of the equation.
[tex]\dfrac{1}{2} \times \dfrac{9}{4} = \dfrac{9}{8};\qquad \left(\dfrac{9}{8}\right)^{2} = \dfrac{81}{64}\\\\x^{2} - \dfrac{9}{4}x + \dfrac{81}{64} = 7\dfrac{81}{64} = \dfrac{529}{64}[/tex]
Step 4. Write the left-hand side as a perfect square
[tex]\dfrac{1}{2} \times \dfrac{9}{4} = \dfrac{9}{8};\qquad \left(\dfrac{9}{8}\right)^{2} = \dfrac{81}{64}\\\\x^{2} - \dfrac{9}{4}x + \dfrac{81}{64} = 7\dfrac{81}{64} = \dfrac{529}{64}[/tex]
Step 5. Take the square root of each side
[tex]x - \dfrac{9}{8} = \pm\sqrt{\dfrac{529}{64}} = \pm\dfrac{23}{8}[/tex]
Step 6. Solve for x
[tex]\begin{array}{rlcrl}x - \dfrac{9}{8} & =\dfrac{23}{8}& \qquad & x - \dfrac{9}{8} & = -\dfrac{23}{8} \\\\x & =\dfrac{23}{8} + \dfrac{9}{8}&\qquad & x & = -\dfrac{23}{8} + \dfrac{9}{8} \\\\x& =\dfrac{32}{8} &\qquad & x & \ -\dfrac{14}{8} \\\\x& =4 & \qquad & x & -1.75 \\\end{array}\\\\\text{f(x) = g(x) when \boxed{\textbf{x = 4 or x = -1.75}}}[/tex]
Check:
[tex]\begin{array}{rlcrl}4^{2} - 2(4) - 8 & = \dfrac{1}{4}(4) -1&\qquad & (-1.75)^{2} - 2(-1.75) - 8 & = \dfrac{1}{4}(-1.75) - 1\\\\16 - 8 -8& = 1 - 1&\qquad & 3.0625 +3.5 - 8 & = -0.4375 - 1 \\\\0& =0&\qquad & -1.4375 & = -1.4375 \\\\\end{array}[/tex]
The diagram below shows that the graph of g(x) intersects that of the parabola ƒ(x) at x = -1.7 and x = 4.
