Mathematics

Question

if f(x)= x squared-2x-8 and g(x)= 1/4x-1 for which values of x is f(x)=g(x)? explain and show work please A. -1.75 and -1.438 B. -1.75 and 4 C. -1.438 D. 4 and 0 ​

2 Answer

  • Answer:

    B. -1.75 and 4

    Step-by-step explanation:

    f(x)=g(x)

    Input the equations

    x²-2x-8 = 1/4x-1

    add 1 to both sides

    and subtract 1/4x from both sides

    x²-2 1/4x - 7 = 0  

    factor

    a + b = -2 1/4

    a * b = -7

    1.75 + -4 = -2 1/4

    1.75 * -4 = -7

    reverse their symbols

    1.75 becomes -1.75 and -4 becomes 4.

  • 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.

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