Sketch the graph of the quadratic function.
f(x) = x2 − 40x − 1,200
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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)	=	

    

y-intercept	y	=	  
Correct: Your answer is correct.
x-intercept	x	=	  
Incorrect: Your answer is incorrect.


To sketch the graph of the quadratic function \( f(x) = x^2 - 40x - 1200 \), let's break it down step by step:

### 1. **Find the Vertex:**
The vertex of a parabola in the form \( f(x) = ax^2 + bx + c \) can be found using the formula:

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

Here, \( a = 1 \), \( b = -40 \), and \( c = -1200 \). Substituting these values into the formula:

\[
x_{\text{vertex}} = -\frac{-40}{2(1)} = \frac{40}{2} = 20
\]

Now, substitute \( x = 20 \) into the original function to find the \( y \)-coordinate of the vertex:

\[
f(20) = (20)^2 - 40(20) - 1200 = 400 - 800 - 1200 = -1600
\]

Thus, the vertex is at \( (20, -1600) \).

### 2. **Find the y-intercept:**
The \( y \)-intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into the function:

\[
f(0) = 0^2 - 40(0) - 1200 = -1200
\]

Thus, the \( y \)-intercept is \( y = -1200 \).

### 3. **Find the x-intercepts:**
The \( x \)-intercepts occur when \( f(x) = 0 \). To find them, solve the equation:

\[
x^2 - 40x - 1200 = 0
\]

We can use the quadratic formula:

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

Substitute \( a = 1 \), \( b = -40 \), and \( c = -1200 \):

\[
x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(-1200)}}{2(1)}
\]
\[
x = \frac{40 \pm \sqrt{1600 + 4800}}{2}
\]
\[
x = \frac{40 \pm \sqrt{6400}}{2}
\]
\[
x = \frac{40 \pm 80}{2}
\]

This gives two solutions:

\[
x = \frac{40 + 80}{2} = \frac{120}{2} = 60
\]
\[
x = \frac{40 - 80}{2} = \frac{-40}{2} = -20
\]

Thus, the \( x \)-intercepts are \( x = -20 \) and \( x = 60 \).

### 4. **Graphing the Function:**
Now, you have all the key points needed to sketch the graph:
- The vertex is at \( (20, -1600) \).
- The \( y \)-intercept is at \( (0, -1200) \).
- The \( x \)-intercepts are at \( x = -20 \) and \( x = 60 \).

The graph will be a parabola opening upwards (since the coefficient of \( x^2 \) is positive), passing through these key points.

### Summary:
- **Vertex**: \( (20, -1600) \)
- **y-intercept**: \( -1200 \)
- **x-intercepts**: \( -20, 60 \)

Would you like me to plot this graph for you?

### Additional Questions:
1. How does changing the coefficient \( a \) affect the shape of the parabola?
2. How would you complete the square to rewrite this quadratic function in vertex form?
3. What is the significance of the discriminant in finding the x-intercepts?
4. What is the axis of symmetry for the parabola?
5. How does the y-intercept relate to the constant term in the quadratic equation?

### Tip:
When graphing quadratics, always identify the vertex, axis of symmetry, and intercepts first to guide your sketch.

Sketch the graph of the quadratic function. f(x) = x^2 − 40x − 1200. Indicate the coordinates of the vertex, the y-intercept, and the x-intercepts (if any).

This tutorial shows how to graph the quadratic function f(x) = x^2 - 40x - 1200 by identifying the vertex, y-intercept, and x-intercepts. Learn step-by-step how to apply the vertex and quadratic formulas to solve this problem, suitable for students in Grades 9-11.

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