Questions: Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. f(x) = x^2 - 6x - 16

Solution Steps

The quadratic function is given by \( f(x) = x^2 - 6x - 16 \). The vertex form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex. To find the vertex, we use the formula \( h = -\frac{b}{2a} \).

For the given function, \( a = 1 \) and \( b = -6 \).

\[ h = -\frac{-6}{2 \times 1} = 3 \]

Substitute \( x = 3 \) back into the function to find \( k \):

\[ k = f(3) = 3^2 - 6 \times 3 - 16 = 9 - 18 - 16 = -25 \]

Thus, the vertex is \((3, -25)\).

Y-intercept:

The y-intercept occurs when \( x = 0 \).

\[ f(0) = 0^2 - 6 \times 0 - 16 = -16 \]

So, the y-intercept is \((0, -16)\).

X-intercepts:

To find the x-intercepts, set \( f(x) = 0 \):

\[ x^2 - 6x - 16 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times (-16)}}{2 \times 1} \]

\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} \]

\[ x = \frac{6 \pm \sqrt{100}}{2} \]

\[ x = \frac{6 \pm 10}{2} \]

The solutions are \( x = 8 \) and \( x = -2 \). Thus, the x-intercepts are \((8, 0)\) and \((-2, 0)\).

The axis of symmetry for a parabola in the form \( f(x) = ax^2 + bx + c \) is the vertical line \( x = h \). From Step 1, we found \( h = 3 \).

Thus, the axis of symmetry is \( x = 3 \).

Final Answer