# Help Identifying x-intercepts of quadratic functions

• Kalzar89
In summary, the task is to identify the vertex and x-intercepts of a quadratic function given by f(x) = 1/2x^2 - 4. The coefficient is clarified to be 1/2. The student is able to find the vertex but needs help factoring the equation to find the x-intercepts. They have attempted to factor it but are having difficulty. However, it is not necessary to factor in order to find the x-intercepts. The solution is obtained by setting the function equal to 0 and solving for x, resulting in x = ±2√2.
Kalzar89

## Homework Statement

Identify the vertex and x-intercept(s) of the quadratic function algebraically.

## Homework Equations

f(x) = 1/2x^2 - 4

1/2 Is the coefficent if its not clear.

## The Attempt at a Solution

Ok I do not need help finding the vertex I have already got that, I need help factoring it out to find the x-intercepts.

Here is my attempt

0 = 1/2x^2 - 4
0 = 1/2 (x^2 - 8)

That is as far as I can factor it, I am terrible at factoring these and I actually have a way easier time factoring trinomials and polynomials.

Could you please explain how I am to factor farther and how am I supposed to solve for the x-intercepts after its all factored. I have even looked all over the internet for a good factoring tutorial guide but I just cannot even understand them.

Thanks

Last edited:
Kalzar89 said:
Ok I do not need help finding the vertex I have already got that, I need help factoring it out to find the x-intercepts.

Here is my attempt

0 = 1/2x^2 - 4
0 = 1/2 (x^2 - 8)

That is as far as I can factor it, I am terrible at factoring these and I actually have a way easier time factoring trinomials and polynomials.

There is no need to worry that much about factoring. You have

$$\frac{1}{2}x^{2}-4=0.$$​

Simply solve for x to obtain the intercept.

I believe that was what he was trying to do! Kalzar89, there is no need to factor.
$(1/2)x^2- 4= 0$
$(1/2)x^2= 4$
$x^2= 8$
and take the square root of both sides.
$x= \pm\sqrt{8}= \pm \sqrt{(4)(2)}= \pm 2\sqrt{2}$

## 1. What is an x-intercept in a quadratic function?

An x-intercept is a point on the graph of a quadratic function where the graph crosses the x-axis. It is the value of x at which the function's output, or y-value, is equal to zero.

## 2. How can I identify the x-intercepts of a quadratic function?

To identify the x-intercepts of a quadratic function, you can set the function equal to zero and solve for x using the quadratic formula or by factoring the function. The resulting values of x are the x-intercepts.

## 3. Can a quadratic function have more than two x-intercepts?

No, a quadratic function can have at most two x-intercepts. This is because a quadratic function is a polynomial of degree 2, meaning it can have at most two solutions when set equal to zero.

## 4. What does the x-intercept represent in a real-world context?

In a real-world context, the x-intercept represents the value of the independent variable at which the dependent variable becomes equal to zero. For example, in a quadratic function that represents the height of a ball thrown in the air, the x-intercepts would represent the times at which the ball hits the ground.

## 5. How do the coefficients of a quadratic function affect the x-intercepts?

The coefficients of a quadratic function affect the x-intercepts in the following ways:

• The coefficient of x^2 determines the direction and shape of the parabola. If it is positive, the parabola opens upwards and the x-intercepts will be positive and negative. If it is negative, the parabola opens downwards and the x-intercepts will be negative and positive.
• The coefficient of x affects the position of the parabola along the x-axis. It does not directly affect the x-intercepts, but it can shift the parabola left or right, thus changing the values of the x-intercepts.
• The constant term affects the y-intercept of the parabola and does not directly affect the x-intercepts, but it can shift the parabola up or down, thus changing the values of the x-intercepts.

• Precalculus Mathematics Homework Help
Replies
6
Views
709
• Precalculus Mathematics Homework Help
Replies
7
Views
1K
• Precalculus Mathematics Homework Help
Replies
12
Views
2K
• Precalculus Mathematics Homework Help
Replies
2
Views
593
• Precalculus Mathematics Homework Help
Replies
1
Views
856
• Precalculus Mathematics Homework Help
Replies
6
Views
3K
• Precalculus Mathematics Homework Help
Replies
2
Views
1K
• Precalculus Mathematics Homework Help
Replies
12
Views
748
• Precalculus Mathematics Homework Help
Replies
8
Views
216
• Precalculus Mathematics Homework Help
Replies
11
Views
1K