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
domain and range of the function.
f(x)=6x-x^2+16

Use the graphing tool to graph the equation. Use the
vertex and one of the intercepts to draw the graph.

The axis of symmetry is blank.
(Type an equation.)

The domain of the function is blank
(Type your answer in interval notation.)

The range of the function is blank.
(Type your answer in interval notation.)

1 answer

To analyze the quadratic function \( f(x) = 6x - x^2 + 16 \), we can first rewrite it in standard form. Rearranging the terms gives us:

\[ f(x) = -x^2 + 6x + 16 \]

Step 1: Find the Vertex

The vertex of a quadratic function in the form \( ax^2 + bx + c \) can be found using the vertex formula:

\[ x = -\frac{b}{2a} \]

Here, \( a = -1 \) and \( b = 6 \). Substituting these values into the formula gives:

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

To find the \( y \)-coordinate of the vertex, plug \( x = 3 \) back into the function:

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

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

Step 2: Find the x-Intercepts

The x-intercepts occur when \( f(x) = 0 \):

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

Rearranging gives:

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

Now we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -6 \), and \( c = -16 \):

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

Calculating the discriminant:

\[ x = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm \sqrt{100}}{2} = \frac{6 \pm 10}{2} \]

This gives us:

\[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-4}{2} = -2 \]

So the x-intercepts are \( (8, 0) \) and \( (-2, 0) \).

Step 3: Find the y-Intercept

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

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

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

Step 4: Axis of Symmetry

The axis of symmetry for the parabola is given by:

\[ x = 3 \]

Step 5: Domain and Range

For a quadratic function, the domain is all real numbers, which in interval notation is:

\[ \text{Domain: } (-\infty, \infty) \]

Now to find the range, since the parabola opens downwards (as \( a = -1 < 0 \)), the maximum value is at the vertex \( y = 25 \) and goes down to negative infinity. Thus, the range is:

\[ \text{Range: } (-\infty, 25] \]

Summary

  • The equation of the parabola's axis of symmetry is: \( x = 3 \)
  • The domain of the function is: \( (-\infty, \infty) \)
  • The range of the function is: \( (-\infty, 25] \)

You can plot the vertex (3, 25), the x-intercepts (8, 0) and (-2, 0), and the y-intercept (0, 16) to graph the parabola, ensuring it opens downwards. If you need a graphing tool to visualize this, you can use online graphing tools like Desmos or GeoGebra.