Find X-Intercepts of y=(x+1¼)²: Perfect Square Intersection

Quadratic Functions with Perfect Square Form

Find the intersection of the function

y=(x+114)2 y=(x+1\frac{1}{4})^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, we substitute Y =0 and solve to find the intersection point with the X-axis
00:13 Extract the root to eliminate the exponent
00:27 Isolate X
00:36 This is the X value at the intersection point, we substitute Y=0 as we did at the point
00:44 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the intersection of the function

y=(x+114)2 y=(x+1\frac{1}{4})^2

With the X

2

Step-by-step solution

To find the intersection of the function y=(x+114)2 y = (x + 1\frac{1}{4})^2 with the x-axis, we set y=0 y = 0 since intersections on the x-axis have a y y -coordinate of zero.

Therefore, our equation becomes:

(x+114)2=0 (x + 1\frac{1}{4})^2 = 0 .

To solve this equation, take the square root of both sides:

x+114=0 x + 1\frac{1}{4} = 0 .

Next, solve for x x by subtracting 114 1\frac{1}{4} from both sides:

x=114 x = -1\frac{1}{4} .

Thus, the intersection point of the function with the x-axis is (114,0)(-1\frac{1}{4}, 0).

After checking the provided answer choices, the correct choice is:

: (114,0) (-1\frac{1}{4},0)

Therefore, the solution to the problem is (114,0)(-1\frac{1}{4}, 0).

3

Final Answer

(114,0) (-1\frac{1}{4},0)

Key Points to Remember

Essential concepts to master this topic
  • X-Intercepts: Set the function equal to zero to find intersection points
  • Technique: Take square root of both sides when (x+a)2=0 (x + a)^2 = 0
  • Check: Substitute x=114 x = -1\frac{1}{4} back: (114+114)2=02=0 (-1\frac{1}{4} + 1\frac{1}{4})^2 = 0^2 = 0

Common Mistakes

Avoid these frequent errors
  • Confusing x-intercepts with y-intercepts
    Don't set x = 0 to find x-intercepts = gives you the y-intercept instead! This gives you the point (0, y) instead of where the graph crosses the x-axis. Always set y = 0 to find x-intercepts.

Practice Quiz

Test your knowledge with interactive questions

Find the intersection of the function

\( y=(x-2)^2 \)

With the X

FAQ

Everything you need to know about this question

Why do I set y = 0 to find x-intercepts?

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X-intercepts are points where the graph crosses the x-axis. At these points, the y-coordinate is always zero. So setting y = 0 helps you find exactly where this happens!

What if I get a negative x-value like -1¼?

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That's completely normal! Negative x-values just mean the intercept is to the left of the origin. The point (114,0) (-1\frac{1}{4}, 0) is still a valid x-intercept.

Why is there only one x-intercept for this parabola?

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This parabola is in perfect square form (x+a)2 (x + a)^2 , which means it just touches the x-axis at one point. It's like the tip of a U-shape sitting on the x-axis.

How do I convert 1¼ to work with it easier?

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You can write 114 1\frac{1}{4} as 54 \frac{5}{4} or 1.25. Use whichever form feels more comfortable for your calculations!

What's the difference between (x,y) and just finding x?

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When finding x-intercepts, you need the complete coordinate pair. You solve for the x-value, but the final answer is (x,0) (x, 0) because y is always 0 at x-intercepts.

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