To sketch the graph of the quadratic function $f(x) = 6x - x^2 + 7$, we need to perform the following steps:

Step 1: Rewrite the function in standard form

The given function is:

$f(x) = -x^2 + 6x + 7$

This is a quadratic function in the form $f(x) = ax^2 + bx + c$, where $a = -1$, $b = 6$, and $c = 7$.

Step 2: Find the vertex of the parabola

The vertex of a parabola given by the equation $f(x) = ax^2 + bx + c$ is located at:

$x_{\text{vertex}} = \frac{-b}{2a}$

Substitute the values of $a = -1$ and $b = 6$:

$x_{\text{vertex}} = \frac{-6}{2(-1)} = \frac{6}{2} = 3$

To find the corresponding $y$-coordinate of the vertex, substitute $x = 3$ into the equation $f(x) = -x^2 + 6x + 7$:

$f(3) = -(3)^2 + 6(3) + 7 = -9 + 18 + 7 = 16$

Thus, the vertex of the parabola is $(3, 16)$.

Step 3: Find the x-intercepts (roots) of the function

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

$-x^2 + 6x + 7 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute $a = -1$, $b = 6$, and $c = 7$:

$x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(7)}}{2(-1)}$ $x = \frac{-6 \pm \sqrt{36 + 28}}{-2}$ $x = \frac{-6 \pm \sqrt{64}}{-2}$ $x = \frac{-6 \pm 8}{-2}$

Thus, we have two solutions:

$x = \frac{-6 + 8}{-2} = \frac{2}{-2} = -1$ $x = \frac{-6 - 8}{-2} = \frac{-14}{-2} = 7$

So, the x-intercepts are $x = -1$ and $x = 7$.

Step 4: Find the y-intercept

The y-intercept occurs when $x = 0$. Substitute $x = 0$ into the equation:

$f(0) = -(0)^2 + 6(0) + 7 = 7$

So, the y-intercept is $(0, 7)$.

Step 5: Axis of symmetry

The axis of symmetry is the vertical line that passes through the vertex. Since the vertex is $(3, 16)$, the equation of the axis of symmetry is:

$x = 3$

Step 6: Sketch the graph

• The vertex is at $(3, 16)$.
• The x-intercepts are at $x = -1$ and $x = 7$.
• The y-intercept is at $(0, 7)$.
• The axis of symmetry is the line $x = 3$.

The parabola opens downward because the coefficient of $x^2$ (i.e., $-1$) is negative.

Step 7: Domain and Range

• The domain of a quadratic function is always all real numbers because the function is defined for all $x$-values. So, the domain is:

$\text{Domain: } (-\infty, \infty)$

• The range of the quadratic function depends on the vertex. Since the parabola opens downward, the maximum value occurs at the vertex, which is $y = 16$. The range is all values less than or equal to 16:

$\text{Range: } (-\infty, 16]$

Summary:

• Equation of the axis of symmetry: $x = 3$
• Domain: $(-\infty, \infty)$
• Range: $(-\infty, 16]$

Do you need more details on any of these steps or further explanations? Here are 5 related questions you can explore next:

1. How can I graph the quadratic function by plotting points?
2. What happens if the coefficient of $x^2$ changes in the equation?
3. How do you derive the axis of symmetry from the general form of a quadratic function?
4. How do you solve a quadratic equation using factoring instead of the quadratic formula?
5. How do you identify the nature of the roots (real, complex) of a quadratic function?

Tip: When graphing a quadratic function, always check for symmetry. The points on the left of the axis of symmetry are a mirror image of those on the right!