Consider The Quadratic Function F ( X ) = 1 5 X 2 − 5 X + 12 F(x) = \frac{1}{5} X^2 - 5x + 12 F ( X ) = 5 1 ​ X 2 − 5 X + 12 . Which Statements Are True About The Function And Its Graph? Select Three Options.A. The Value Of F ( − 10 ) = 82 F(-10) = 82 F ( − 10 ) = 82 . B. The Graph Of The Function Is A Parabola. C. The

Introduction

A quadratic function is a polynomial function of degree two, which means the highest power of the variable is two. The general form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. In this article, we will consider the quadratic function $f(x) = \frac{1}{5} x^2 - 5x + 12$ and examine its properties and graph.

Properties of the Quadratic Function

Coefficients of the Quadratic Function

The given quadratic function is $f(x) = \frac{1}{5} x^2 - 5x + 12$. The coefficients of this function are $a = \frac{1}{5}$, $b = -5$, and $c = 12$.

Determining the Value of $f(-10)$

To determine the value of $f(-10)$, we substitute $x = -10$ into the function:

$f(-10) = \frac{1}{5} (-10)^2 - 5(-10) + 12$

$f(-10) = \frac{1}{5} (100) + 50 + 12$

$f(-10) = 20 + 50 + 12$

$f(-10) = 82$

Therefore, the value of $f(-10)$ is indeed 82, making statement A true.

Graph of the Quadratic Function

The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that opens upwards or downwards. The direction of the opening of the parabola depends on the sign of the coefficient $a$. If $a$ is positive, the parabola opens upwards, and if $a$ is negative, the parabola opens downwards.

In this case, the coefficient $a$ is $\frac{1}{5}$, which is positive. Therefore, the graph of the function $f(x) = \frac{1}{5} x^2 - 5x + 12$ is a parabola that opens upwards.

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that passes through the vertex of the parabola. The vertex of a parabola is the point where the parabola changes direction.

To find the axis of symmetry, we need to find the x-coordinate of the vertex. The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$.

In this case, $a = \frac{1}{5}$ and $b = -5$. Therefore, the x-coordinate of the vertex is:

$x = -\frac{-5}{2(\frac{1}{5})}$

$x = -\frac{-5}{\frac{2}{5}}$

$x = -\frac{-5}{\frac{2}{5}} \times \frac{5}{5}$

$x = \frac{-5 \times 5}{2}$

$x = -\frac{25}{2}$

Therefore, the axis of symmetry is the vertical line $x = -\frac{25}{2}$.

Vertex of the Parabola

The vertex of a parabola is the point where the parabola changes direction. The vertex is given by the formula $(x, y) = (-\frac{b}{2a}, f(-\frac{b}{2a}))$.

In this case, $a = \frac{1}{5}$, $b = -5$, and $c = 12$. Therefore, the vertex is:

$(x, y) = (-\frac{-5}{2(\frac{1}{5})}, f(-\frac{-5}{2(\frac{1}{5})}))$

$(x, y) = (-\frac{-5}{\frac{2}{5}}, f(-\frac{-5}{\frac{2}{5}}))$

$(x, y) = (-\frac{-5}{\frac{2}{5}} \times \frac{5}{5}, f(-\frac{-5}{\frac{2}{5}} \times \frac{5}{5}))$

$(x, y) = (-\frac{-5 \times 5}{2}, f(-\frac{-5 \times 5}{2}))$

$(x, y) = (-\frac{25}{2}, f(-\frac{25}{2}))$

To find the y-coordinate of the vertex, we substitute $x = -\frac{25}{2}$ into the function:

$f(-\frac{25}{2}) = \frac{1}{5} (-\frac{25}{2})^2 - 5(-\frac{25}{2}) + 12$

$f(-\frac{25}{2}) = \frac{1}{5} (\frac{625}{4}) + \frac{125}{2} + 12$

$f(-\frac{25}{2}) = \frac{625}{20} + \frac{125}{2} + 12$

$f(-\frac{25}{2}) = 31.25 + 62.5 + 12$

$f(-\frac{25}{2}) = 105.75$

Therefore, the vertex of the parabola is the point $(-\frac{25}{2}, 105.75)$.

Intercepts of the Parabola

The x-intercepts of a parabola are the points where the parabola intersects the x-axis. The y-intercept of a parabola is the point where the parabola intersects the y-axis.

To find the x-intercepts, we set $y = 0$ and solve for $x$:

$\frac{1}{5} x^2 - 5x + 12 = 0$

Multiplying both sides by 5:

$x^2 - 25x + 60 = 0$

Factoring the quadratic:

$(x - 20)(x - 3) = 0$

Solving for $x$:

$x - 20 = 0$ or $x - 3 = 0$

$x = 20$ or $x = 3$

Therefore, the x-intercepts are the points $(20, 0)$ and $(3, 0)$.

To find the y-intercept, we set $x = 0$ and solve for $y$:

$f(0) = \frac{1}{5} (0)^2 - 5(0) + 12$

$f(0) = 0 - 0 + 12$

$f(0) = 12$

Therefore, the y-intercept is the point $(0, 12)$.

Conclusion

In conclusion, the quadratic function $f(x) = \frac{1}{5} x^2 - 5x + 12$ has the following properties:

• The value of $f(-10)$ is 82.
• The graph of the function is a parabola that opens upwards.
• The axis of symmetry is the vertical line $x = -\frac{25}{2}$.
• The vertex of the parabola is the point $(-\frac{25}{2}, 105.75)$.
• The x-intercepts are the points $(20, 0)$ and $(3, 0)$.
• The y-intercept is the point $(0, 12)$.

Therefore, statements A, B, and C are all true.

Answer Key

A. The value of $f(-10) = 82$. True

B. The graph of the function is a parabola. True

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Introduction

In our previous article, we explored the properties of the quadratic function $f(x) = \frac{1}{5} x^2 - 5x + 12$. In this article, we will answer some frequently asked questions about quadratic functions and their graphs.

Q&A

Q: What is the general form of a quadratic function?

A: The general form of a quadratic function is $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.

Q: What is the axis of symmetry of a parabola?

A: The axis of symmetry of a parabola is a vertical line that passes through the vertex of the parabola. The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$.

Q: How do I find the vertex of a parabola?

A: To find the vertex of a parabola, you need to find the x-coordinate of the vertex using the formula $x = -\frac{b}{2a}$. Then, substitute this value of $x$ into the function to find the y-coordinate of the vertex.

Q: What are the x-intercepts of a parabola?

A: The x-intercepts of a parabola are the points where the parabola intersects the x-axis. To find the x-intercepts, set $y = 0$ and solve for $x$.

Q: What is the y-intercept of a parabola?

A: The y-intercept of a parabola is the point where the parabola intersects the y-axis. To find the y-intercept, set $x = 0$ and solve for $y$.

Q: How do I determine the direction of the opening of a parabola?

A: To determine the direction of the opening of a parabola, look at the sign of the coefficient $a$. If $a$ is positive, the parabola opens upwards. If $a$ is negative, the parabola opens downwards.

Q: Can a quadratic function have more than one x-intercept?

A: Yes, a quadratic function can have more than one x-intercept. This occurs when the quadratic function has two real roots.

Q: Can a quadratic function have no x-intercepts?

A: Yes, a quadratic function can have no x-intercepts. This occurs when the quadratic function has no real roots.

Q: Can a quadratic function have more than one y-intercept?

A: No, a quadratic function can have only one y-intercept.

Q: Can a quadratic function have no y-intercept?

A: Yes, a quadratic function can have no y-intercept. This occurs when the quadratic function has a negative leading coefficient.

Conclusion

In conclusion, quadratic functions and their graphs have many interesting properties. By understanding these properties, you can better analyze and solve problems involving quadratic functions.

Additional Resources

For more information on quadratic functions and their graphs, check out the following resources:

• Khan Academy: Quadratic Functions
• Mathway: Quadratic Functions
• Wolfram Alpha: Quadratic Functions

Practice Problems

Try the following practice problems to test your understanding of quadratic functions and their graphs:

1. Find the x-intercepts of the quadratic function $f(x) = x^2 - 6x + 8$.
2. Find the y-intercept of the quadratic function $f(x) = 2x^2 + 3x - 1$.
3. Determine the direction of the opening of the parabola given by the quadratic function $f(x) = -x^2 + 4x - 3$.

Answer Key

1. The x-intercepts are the points $(4, 0)$ and $(2, 0)$.
2. The y-intercept is the point $(0, -1)$.
3. The parabola opens downwards.