Finding X-Intercepts of y=(x-1/2)²: Quadratic Function Analysis

Q: Why does this parabola have only one x-intercept?

This parabola is a perfect square that just touches the x-axis at one point. The vertex $ (\frac{1}{2}, 0) $ is also the x-intercept, making it a double root.

Q: How do I know when y = 0 at the x-axis?

The x-axis is defined as all points where y = 0. So to find x-intercepts, you always set the function equal to zero and solve for x values.

Q: What if I expand the squared term first?

You could expand $ (x - \frac{1}{2})^2 $ to get $ x^2 - x + \frac{1}{4} $, but it's much easier to work with the factored form when finding intercepts!

Q: Is the answer a point or just a number?

X-intercepts are points with coordinates $ (x, 0) $. Since we found x = 1/2 and y = 0 at the intercept, the complete answer is the point $ (\frac{1}{2}, 0) $.

Q: Why can't (0, 0) be the x-intercept?

Let's check: when x = 0, $ y = (0 - \frac{1}{2})^2 = (-\frac{1}{2})^2 = \frac{1}{4} $. Since y ≠ 0 when x = 0, the point (0, 0) is not on this parabola.

X-Intercepts with Perfect Square Form

Find the intersection of the function

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

With the X

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the intersection point of the function with the X-axis

00:03 At the intersection point with the X-axis, Y = 0

00:07 Therefore, substitute Y = 0 and solve to find the intersection point with the X-axis

00:11 Take the square root to eliminate the power

00:24 Isolate X

00:34 This is the X value at the intersection point, substitute Y=0 as we did at the point

00:40 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the intersection of the function

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

With the X

Step-by-step solution

To find the intersection of the parabola $y = (x - \frac{1}{2})^2$ with the x-axis, we need to set $y = 0$ because at the x-axis, the y-coordinate is always zero.

Let's go through the steps:

Step 1: Set the equation equal to zero: $(x - \frac{1}{2})^2 = 0$ .
Step 2: Solve for $x$ . A square equals zero only when the base itself is zero, thus:
$x - \frac{1}{2} = 0$
Step 3: Solve this equation for $x$ :
$x = \frac{1}{2}$

The intersection point on the x-axis has coordinates $(x, y)$ , where we have found $x = \frac{1}{2}$ and we know $y = 0$ .

Therefore, the intersection of the function with the x-axis is at the point $(\frac{1}{2}, 0)$ .

Thus, the correct answer is choice 3: $(\frac{1}{2}, 0)$ .

Final Answer

$(\frac{1}{2},0)$

Key Points to Remember

Essential concepts to master this topic

X-Intercept Definition: Set y = 0 and solve for x coordinates
Perfect Square Method: If $(x - \frac{1}{2})^2 = 0$ , then $x - \frac{1}{2} = 0$
Verification: Check that $y = (\frac{1}{2} - \frac{1}{2})^2 = 0^2 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Confusing x-intercepts with vertex coordinates
Don't think the vertex at $(\frac{1}{2}, 0)$ means it's not an x-intercept! Students often assume a parabola can't touch the x-axis at its vertex. Always remember x-intercepts occur when y = 0, regardless of whether it's also the vertex.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

$$ y=(x+4)^2 $$

With the Y

FAQ

Everything you need to know about this question

Why does this parabola have only one x-intercept?

This parabola is a perfect square that just touches the x-axis at one point. The vertex $(\frac{1}{2}, 0)$ is also the x-intercept, making it a double root.

How do I know when y = 0 at the x-axis?

The x-axis is defined as all points where y = 0. So to find x-intercepts, you always set the function equal to zero and solve for x values.

What if I expand the squared term first?

You could expand $(x - \frac{1}{2})^2$ to get $x^2 - x + \frac{1}{4}$ , but it's much easier to work with the factored form when finding intercepts!

Is the answer a point or just a number?

X-intercepts are points with coordinates $(x, 0)$ . Since we found x = 1/2 and y = 0 at the intercept, the complete answer is the point $(\frac{1}{2}, 0)$ .

Why can't (0, 0) be the x-intercept?

Let's check: when x = 0, $y = (0 - \frac{1}{2})^2 = (-\frac{1}{2})^2 = \frac{1}{4}$ . Since y ≠ 0 when x = 0, the point (0, 0) is not on this parabola.

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Finding X-Intercepts of y=(x-1/2)²: Quadratic Function Analysis

X-Intercepts with Perfect Square Form

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why does this parabola have only one x-intercept?

How do I know when y = 0 at the x-axis?

What if I expand the squared term first?

Is the answer a point or just a number?

Why can't (0, 0) be the x-intercept?

🌟 Unlock Your Math Potential

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