(a) Draw The Graph Of The Quadratic Function:${ \left{(x, Y) \mid Y = X^2 - 2x - 3\right} }$for The Domain { -2 \leq X \leq 4$}$.(b) Estimate The Gradients Of The Curve At The Points Where { X = 1$}$ And [$x =

Introduction

Quadratic functions are a fundamental concept in mathematics, and their graphical representation is a crucial aspect of understanding these functions. In this article, we will delve into the world of quadratic functions and explore the graph of the function $y = x^2 - 2x - 3$ for the domain $-2 \leq x \leq 4$. We will also estimate the gradients of the curve at the points where $x = 1$ and $x = 3$.

What are Quadratic Functions?

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. The graph of a quadratic function is a parabola, which is a U-shaped curve that opens upwards or downwards.

The Graph of the Quadratic Function

The given quadratic function is $y = x^2 - 2x - 3$. To draw the graph of this function, we need to find the x-intercepts, the vertex, and the y-intercept.

Finding the x-Intercepts

The x-intercepts are the points where the graph of the function crosses the x-axis. To find the x-intercepts, we need to set $y = 0$ and solve for $x$.

$$0 = x^2 - 2x - 3$$

We can factor the quadratic expression as:

$$(x - 3)(x + 1) = 0$$

This gives us two possible values for $x$: $x = 3$ and $x = -1$. Therefore, the x-intercepts are $(3, 0)$ and $(-1, 0)$.

Finding the Vertex

The vertex of a parabola is the point where the graph changes direction. To find the vertex, we need to use the formula:

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

In this case, $a = 1$ and $b = -2$. Plugging these values into the formula, we get:

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

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by plugging $x = 1$ into the original equation:

$$y = (1)^2 - 2(1) - 3 = -4$$

Therefore, the vertex is $(1, -4)$.

Finding the y-Intercept

The y-intercept is the point where the graph of the function crosses the y-axis. To find the y-intercept, we need to set $x = 0$ and solve for $y$.

$$y = (0)^2 - 2(0) - 3 = -3$$

Therefore, the y-intercept is $(0, -3)$.

Drawing the Graph

Now that we have found the x-intercepts, the vertex, and the y-intercept, we can draw the graph of the quadratic function.

The graph of the function $y = x^2 - 2x - 3$ is a parabola that opens upwards. The x-intercepts are $(3, 0)$ and $(-1, 0)$, the vertex is $(1, -4)$, and the y-intercept is $(0, -3)$.

Estimating the Gradients

The gradient of a curve at a point is the rate of change of the function at that point. To estimate the gradients of the curve at the points where $x = 1$ and $x = 3$, we need to use the formula:

$$\text{Gradient} = \frac{\text{change in } y}{\text{change in } x}$$

Estimating the Gradient at $x = 1$

To estimate the gradient at $x = 1$, we need to find the change in $y$ and the change in $x$. We can do this by plugging $x = 1$ into the original equation and finding the corresponding value of $y$.

$$y = (1)^2 - 2(1) - 3 = -4$$

Now that we have the value of $y$, we can find the change in $y$ by subtracting the value of $y$ at $x = 0$ from the value of $y$ at $x = 1$.

$$\text{change in } y = -4 - (-3) = -1$$

We can also find the change in $x$ by subtracting the value of $x$ at $x = 0$ from the value of $x$ at $x = 1$.

$$\text{change in } x = 1 - 0 = 1$$

Now that we have the change in $y$ and the change in $x$, we can estimate the gradient at $x = 1$.

$$\text{Gradient} = \frac{\text{change in } y}{\text{change in } x} = \frac{-1}{1} = -1$$

Estimating the Gradient at $x = 3$

To estimate the gradient at $x = 3$, we need to find the change in $y$ and the change in $x$. We can do this by plugging $x = 3$ into the original equation and finding the corresponding value of $y$.

$$y = (3)^2 - 2(3) - 3 = 0$$

Now that we have the value of $y$, we can find the change in $y$ by subtracting the value of $y$ at $x = 2$ from the value of $y$ at $x = 3$.

$$\text{change in } y = 0 - 3 = -3$$

We can also find the change in $x$ by subtracting the value of $x$ at $x = 2$ from the value of $x$ at $x = 3$.

$$\text{change in } x = 3 - 2 = 1$$

Now that we have the change in $y$ and the change in $x$, we can estimate the gradient at $x = 3$.

$$\text{Gradient} = \frac{\text{change in } y}{\text{change in } x} = \frac{-3}{1} = -3$$

Conclusion

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Frequently Asked Questions

In this article, we will answer some of the most frequently asked questions about quadratic functions.

Q: What is a quadratic function?

A: 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.

Q: What is the graph of a quadratic function?

A: The graph of a quadratic function is a parabola, which is a U-shaped curve that opens upwards or downwards.

Q: How do I find the x-intercepts of a quadratic function?

A: To find the x-intercepts of a quadratic function, you need to set $y = 0$ and solve for $x$. This will give you the points where the graph of the function crosses the x-axis.

Q: How do I find the vertex of a quadratic function?

A: To find the vertex of a quadratic function, you need to use the formula:

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

This will give you the x-coordinate of the vertex. To find the y-coordinate, you need to plug the x-coordinate into the original equation.

Q: How do I find the y-intercept of a quadratic function?

A: To find the y-intercept of a quadratic function, you need to set $x = 0$ and solve for $y$. This will give you the point where the graph of the function crosses the y-axis.

Q: What is the gradient of a quadratic function?

A: The gradient of a quadratic function is the rate of change of the function at a given point. It can be estimated using the formula:

$$\text{Gradient} = \frac{\text{change in } y}{\text{change in } x}$$

Q: How do I estimate the gradient of a quadratic function?

A: To estimate the gradient of a quadratic function, you need to find the change in $y$ and the change in $x$. You can do this by plugging the x-coordinate into the original equation and finding the corresponding value of $y$. Then, you can find the change in $y$ by subtracting the value of $y$ at the previous x-coordinate from the value of $y$ at the current x-coordinate. Similarly, you can find the change in $x$ by subtracting the previous x-coordinate from the current x-coordinate.

Q: What is the difference between a quadratic function and a linear function?

A: A quadratic function is a polynomial function of degree two, while a linear function is a polynomial function of degree one. The graph of a quadratic function is a parabola, while the graph of a linear function is a straight line.

Q: Can I use a quadratic function to model real-world situations?

A: Yes, quadratic functions can be used to model real-world situations. For example, the height of a projectile as a function of time can be modeled using a quadratic function.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you need to set the equation equal to zero and solve for $x$. This can be done using factoring, the quadratic formula, or other methods.

Conclusion

In this article, we have answered some of the most frequently asked questions about quadratic functions. We hope that this article has been helpful in understanding quadratic functions and their applications.