Sketch the graph of the quadratic function. (Select Update Graph to see your response plotted on the screen. Select the Submit button to grade your response.) f(x) = x2 + x − 7 Indicate the coordinates of the vertex, the y-intercept, and the x-intercepts (if any). (If an answer does not exist, enter DNE. If there are multiple x-intercepts, enter your answers as a comma-separated list.) vertex (x, y)

Let's break this down step by step for the quadratic function:

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

1. Vertex:

The vertex of a quadratic function in the form $f(x) = ax^2 + bx + c$ can be found using the formula for the x-coordinate of the vertex:

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

Here, $a = 1$ and $b = 1$. So,

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

Now, substitute $x = -\frac{1}{2}$ into the equation to find the y-coordinate:

$f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 7$

$f\left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 7 = -\frac{27}{4}$

Thus, the vertex is at:

$\left(-\frac{1}{2}, -\frac{27}{4}\right)$

2. y-intercept:

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

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

So, the y-intercept is at:

$(0, -7)$

3. x-intercepts:

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

$x^2 + x - 7 = 0$

Use the quadratic formula:

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

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

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-7)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2}$

So, the x-intercepts are:

$x = \frac{-1 + \sqrt{29}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{29}}{2}$

These are approximate values:

$x \approx 2.19 \quad \text{and} \quad x \approx -3.19$

4. Additional point using symmetry:

The graph is symmetric around the vertical line $x = -\frac{1}{2}$. Since we know the y-intercept is at $(0, -7)$, reflect this point across $x = -\frac{1}{2}$. The distance between $x = 0$ and $x = -\frac{1}{2}$ is $\frac{1}{2}$, so we move $\frac{1}{2}$ units to the left:

$x = -1$

Now, substitute $x = -1$ into the function to find the corresponding y-value:

$f(-1) = (-1)^2 + (-1) - 7 = 1 - 1 - 7 = -7$

So, the additional point is:

$(-1, -7)$

5. Left-hand and right-hand behavior:

For large positive or negative $x$, the quadratic term $x^2$ dominates the behavior of the function.

• As $x \to -\infty$, $f(x) \to +\infty$.
• As $x \to +\infty$, $f(x) \to +\infty$.

Summary of Key Points:

• Vertex: $\left(-\frac{1}{2}, -\frac{27}{4}\right)$
• y-intercept: $(0, -7)$
• x-intercepts: $\frac{-1 + \sqrt{29}}{2}, \frac{-1 - \sqrt{29}}{2} \approx 2.19, -3.19$
• Additional Point: $(-1, -7)$
• Left-hand behavior: As $x \to -\infty$, $f(x) \to +\infty$.
• Right-hand behavior: As $x \to +\infty$, $f(x) \to +\infty$.

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Would you like me to go over any specific part in more detail?
Here are some related questions:

1. How do you find the axis of symmetry for a quadratic function?
2. What determines the direction of a parabola's opening?
3. How do you use the discriminant to find the number of x-intercepts?
4. How do you complete the square to find the vertex form of a quadratic function?
5. Can the quadratic formula ever give complex solutions?

Tip: The coefficient of the quadratic term ($a$) determines whether the parabola opens upwards ($a > 0$) or downwards ($a < 0$).