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Graph the following quadratic functions, determine their domain and range: (a) y = -3x2 (b) y = -2x2 - 8 (c) y = x2 − 9 for $x ",

Mathematics

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Part 1· 6 steps

1 (a) y=−3x2y = -3x^2

  1. Create a table of values for x∈{−2,−1,0,1,2}x \in \{-2, -1, 0, 1, 2\}.

    xy=−3x2−2−3(−2)2=−12−1−3(−1)2=−30−3(0)2=01−3(1)2=−32−3(2)2=−12\begin{array}{|c|c|} \hline x & y = -3x^2 \\ \hline -2 & -3(-2)^2 = -12 \\ -1 & -3(-1)^2 = -3 \\ 0 & -3(0)^2 = 0 \\ 1 & -3(1)^2 = -3 \\ 2 & -3(2)^2 = -12 \\ \hline \end{array}

  2. Describe the graph, domain, and range.

    The graph is a parabola opening downwards with its vertex at (0,0)(0,0).
    Domain: x∈R\boxed{x \in \mathbb{R}}
    Range: y≤0\boxed{y \le 0}

    1 (b) y=−2x2−8y = -2x^2 - 8

  3. Create a table of values for x∈{−2,−1,0,1,2}x \in \{-2, -1, 0, 1, 2\}.

    xy=−2x2−8−2−2(−2)2−8=−16−1−2(−1)2−8=−100−2(0)2−8=−81−2(1)2−8=−102−2(2)2−8=−16\begin{array}{|c|c|} \hline x & y = -2x^2 - 8 \\ \hline -2 & -2(-2)^2 - 8 = -16 \\ -1 & -2(-1)^2 - 8 = -10 \\ 0 & -2(0)^2 - 8 = -8 \\ 1 & -2(1)^2 - 8 = -10 \\ 2 & -2(2)^2 - 8 = -16 \\ \hline \end{array}

  4. Describe the graph, domain, and range.

    The graph is a parabola opening downwards with its vertex at (0,−8)(0,-8).
    Domain: x∈R\boxed{x \in \mathbb{R}}
    Range: y≤−8\boxed{y \le -8}

    1 (c) y=x2−9y = x^2 - 9

  5. Create a table of values for x∈{−2,−1,0,1,2}x \in \{-2, -1, 0, 1, 2\}.

    xy=x2−9−2(−2)2−9=−5−1(−1)2−9=−80(0)2−9=−91(1)2−9=−82(2)2−9=−5\begin{array}{|c|c|} \hline x & y = x^2 - 9 \\ \hline -2 & (-2)^2 - 9 = -5 \\ -1 & (-1)^2 - 9 = -8 \\ 0 & (0)^2 - 9 = -9 \\ 1 & (1)^2 - 9 = -8 \\ 2 & (2)^2 - 9 = -5 \\ \hline \end{array}

  6. Describe the graph, domain, and range.

    The graph is a parabola opening upwards with its vertex at (0,−9)(0,-9).
    Domain: x∈R\boxed{x \in \mathbb{R}}
    Range: y≥−9\boxed{y \ge -9}

Part 2· 4 steps

Given: y=4x2−8y = 4x^2 - 8

2 (a) Shape of the graph
The graph is a parabola. Since the coefficient of x2x^2 is positive (4>04 > 0), it opens upwards.
Shape: Parabola opening upwards\boxed{\text{Parabola opening upwards}}

2 (b) x-intercept(s)

  1. Set y=0y = 0 to find the x-intercepts.

    0=4x2−80 = 4x^2 - 8

  2. Solve for xx.

    4x2=84x^2 = 8 x2=84x^2 = \frac{8}{4} x2=2x^2 = 2 x=±2x = \pm \sqrt{2}

    x-intercepts: (2,0) and (−2,0)\boxed{(\sqrt{2}, 0) \text{ and } (-\sqrt{2}, 0)}

    2 (c) y-intercept

  3. Set x=0x = 0 to find the y-intercept.

    y=4(0)2−8y = 4(0)^2 - 8

  4. Calculate yy.

    y=−8y = -8

    y-intercept: (0,−8)\boxed{(0, -8)}

    2 (d) Turning point
    For a quadratic function of the form y=ax2+cy = ax^2 + c, the turning point is (0,c)(0, c).
    Turning point: (0,−8)\boxed{(0, -8)}

    2 (e) Domain
    For any quadratic function, the domain is all real numbers.
    Domain: x∈R\boxed{x \in \mathbb{R}}

    2 (f) Range
    Since the parabola opens upwards and its turning point (minimum value) is at y=−8y = -8.
    Range: y≥−8\boxed{y \ge -8}

    2 (g) Is the graph a function or a non-function?
    The graph of y=4x2−8y = 4x^2 - 8 is a parabola that passes the vertical line test.
    It is a function\boxed{\text{function}}.

Part 3· 6 steps

Calculate the roots of:

3 (a) y=3x2y = 3x^2

  1. Set y=0y = 0 to find the roots.

    0=3x20 = 3x^2

  2. Solve for xx.

    x2=03x^2 = \frac{0}{3} x2=0x^2 = 0 x=0x = 0

    Roots: x=0\boxed{x = 0}

    3 (b) y=−x2+4y = -x^2 + 4

  3. Set y=0y = 0 to find the roots.

    0=−x2+40 = -x^2 + 4

  4. Solve for xx.

    x2=4x^2 = 4 x=±4x = \pm \sqrt{4} x=±2x = \pm 2

    Roots: x=2 and x=−2\boxed{x = 2 \text{ and } x = -2}

    3 (c) y=2x2+8y = 2x^2 + 8

  5. Set y=0y = 0 to find the roots.

    0=2x2+80 = 2x^2 + 8

  6. Solve for xx.

    2x2=−82x^2 = -8 x2=−82x^2 = \frac{-8}{2} x2=−4x^2 = -4

    Since the square of a real number cannot be negative, there are no real roots.
    Roots: No real roots\boxed{\text{No real roots}}

Part 4· 3 steps

Given: f(x)=−3x2+12f(x) = -3x^2 + 12

4 (a) Draw a sketch of f(x). Clearly indicate the coordinates of the intercepts on the axes as well as the turning point.

  1. Determine the shape and turning point.

    The function is a parabola f(x)=ax2+cf(x) = ax^2 + c with a=−3a = -3 and c=12c = 12.
    Since a=−3<0a = -3 < 0, the parabola opens downwards.
    The turning point is (0,c)(0, c), so it is (0,12)(0, 12).

  2. Determine the y-intercept.

    Set x=0x = 0:

    f(0)=−3(0)2+12=12f(0) = -3(0)^2 + 12 = 12

    The y-intercept is (0,12)(0, 12).

  3. Determine the x-intercepts.

    Set f(x)=0f(x) = 0:

    0=−3x2+120 = -3x^2 + 12 3x2=123x^2 = 12 x2=123x^2 = \frac{12}{3} x2=4x^2 = 4 x=±4x = \pm \sqrt{4} x=±2x = \pm 2

    The x-intercepts are (−2,0)(-2, 0) and (2,0)(2, 0).

    Sketch description:
    The graph is a parabola opening downwards.
    It passes through the x-axis at (−2,0)(-2, 0) and (2,0)(2, 0).
    It passes through the y-axis at (0,12)(0, 12).
    The turning point (which is a maximum) is at (0,12)(0, 12).

    4 (b) Does f(x) have a maximum or minimum value? Determine this value.
    Since the parabola opens downwards (a=−3<0a = -3 < 0), it has a maximum value.
    The maximum value is the y-coordinate of the turning point.
    Maximum value: y=12\boxed{y = 12}

    **4 (c) What is the domain and range of f(

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