Find X-Axis Intersections of y = x² + 4: Quadratic Function Analysis

Q: Does this mean the parabola doesn't exist?

No! The parabola $ y = x^2 + 4 $ definitely exists - it's just shifted 4 units up from $ y = x^2 $, so it never touches the x-axis.

Q: How can I tell if a parabola has x-intercepts without graphing?

Look at the constant term! If your parabola is $ y = x^2 + c $ and c is positive, it has no x-intercepts. If c is negative, it has two x-intercepts.

Q: Where does this parabola cross the y-axis?

To find the y-intercept, set x = 0: $ y = 0^2 + 4 = 4 $. So it crosses the y-axis at (0, 4).

Quadratic Functions with No X-Intercepts

Determine the points of intersection of the function

$y=x^2+4$

With the X

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:07 Let's find where the graph meets the X-axis.

00:11 Remember, at this point, Y equals zero.

00:14 So, we replace Y with zero in our equation. Let's solve for X.

00:20 We need to get X by itself. Let's isolate it.

00:26 Any number squared is always positive, right?

00:47 This means the graph doesn't touch the X-axis.

00:51 And that's the answer to our problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Determine the points of intersection of the function

$y=x^2+4$

With the X

Step-by-step solution

To determine the intersection points of the function $y = x^2 + 4$ with the x-axis, we set the equation to zero, i.e., find where $x^2 + 4 = 0$ .

Let's solve the equation:

$x^2 + 4 = 0$
Subtract 4 on both sides: $x^2 = -4$
Since $x^2 = -4$ has no real solutions (the square of a real number cannot be negative), there are no real $x$ -intercepts.

Therefore, the parabola defined by $y = x^2 + 4$ does not intersect the x-axis.

The solution to the problem is No solution.

Final Answer

No solution

Key Points to Remember

Essential concepts to master this topic

X-Intercepts: Set y = 0 and solve the resulting equation
Technique: From $x^2 + 4 = 0$ get $x^2 = -4$
Check: Since squares of real numbers are never negative, no solution ✓

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 (0,4)! X-intercepts occur where the graph crosses the x-axis, so y must equal zero. Always set y = 0 and solve for x to find x-intercepts.

Practice Quiz

Test your knowledge with interactive questions

Which chart represents the function $$ y=x^2-9 $$ ?

FAQ

Everything you need to know about this question

Why can't x² equal -4?

Because when you square any real number, positive or negative, the result is always positive or zero. For example: $3^2 = 9$ and $(-3)^2 = 9$ , never negative!

Does this mean the parabola doesn't exist?

No! The parabola $y = x^2 + 4$ definitely exists - it's just shifted 4 units up from $y = x^2$ , so it never touches the x-axis.

How can I tell if a parabola has x-intercepts without graphing?

Look at the constant term! If your parabola is $y = x^2 + c$ and c is positive, it has no x-intercepts. If c is negative, it has two x-intercepts.

What's the difference between 'no solution' and 'no real solution'?

Great question! In this context, they mean the same thing. We say 'no solution' because we're only working with real numbers in basic algebra.

Where does this parabola cross the y-axis?

To find the y-intercept, set x = 0: $y = 0^2 + 4 = 4$ . So it crosses the y-axis at (0, 4).

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