Let's solve this step by step!

Part 1: Graphing the Quadratic Equation

We are given the quadratic function:

$f(x) = x^2 - 2x - 15$

Step 1: Find the vertex.

To find the vertex of a parabola in the form $f(x) = ax^2 + bx + c$, we use the formula for the x-coordinate of the vertex:

$x = \frac{-b}{2a}$

For our function $f(x) = x^2 - 2x - 15$, we have $a = 1$ and $b = -2$. Substituting into the formula:

$x = \frac{-(-2)}{2(1)} = \frac{2}{2} = 1$

Now, substitute $x = 1$ into the original function to find the y-coordinate of the vertex:

$f(1) = (1)^2 - 2(1) - 15 = 1 - 2 - 15 = -16$

So, the vertex is $(1, -16)$.

Step 2: Find the intercepts.

• x-intercepts: Set $f(x) = 0$ and solve for $x$.

$0 = x^2 - 2x - 15$

We can solve this quadratic equation by factoring or using the quadratic formula. First, let's try factoring:

$x^2 - 2x - 15 = (x - 5)(x + 3)$

So, the x-intercepts are $x = 5$ and $x = -3$, giving the points $(5, 0)$ and $(-3, 0)$.

• y-intercept: Set $x = 0$ in the function to find $f(0)$:

$f(0) = (0)^2 - 2(0) - 15 = -15$

So, the y-intercept is $(0, -15)$.

Part 2: Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that passes through the vertex. Since the vertex is $(1, -16)$, the axis of symmetry is the vertical line that passes through $x = 1$.

So, the equation of the axis of symmetry is:

$x = 1$

Part 3: Domain of $f(x)$

For a quadratic function, the domain is always all real numbers since there are no restrictions on the input values.

Thus, the domain of $f(x)$ is:

$(-\infty, \infty)$

Part 4: Range of $f(x)$

Since the parabola opens upwards (the coefficient of $x^2$ is positive), the range is determined by the vertex's y-coordinate, which is $-16$. The function's values increase as $x$ moves away from 1 in either direction.

Thus, the range of $f(x)$ is:

$[-16, \infty)$

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Summary of Answers:

• Axis of Symmetry: $x = 1$
• Domain: $(-\infty, \infty)$
• Range: $[-16, \infty)$

Let me know if you'd like to go over any part in more detail or have additional questions!

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5 Related Questions:

1. How can we graph a quadratic function when given the vertex and one point?
2. How do we identify whether a parabola opens upwards or downwards?
3. What is the quadratic formula, and when do we use it?
4. How do we find the vertex of a quadratic equation that is not easily factorable?
5. What does the discriminant tell us about the number of real solutions for a quadratic equation?

Tip:

To quickly find the vertex of any quadratic function, always use the formula $x = \frac{-b}{2a}$ and plug it back into the equation to get the y-coordinate. This is faster than trying to complete the square!