Find X-Axis Intersections of y=(x+3)²+4: Quadratic Function Analysis

Q: Why doesn't this parabola touch the x-axis?

This parabola opens upward with its vertex at (-3, 4). Since the lowest point is 4 units above the x-axis, the entire parabola stays above y = 0 and never intersects it.

Q: How can I tell without solving if there are x-intercepts?

Look at the vertex form $ y = (x+3)^2 + 4 $. The "+4" tells you the vertex is 4 units above the x-axis. Since parabolas with positive leading coefficients open upward, this one never goes below y = 4.

Q: What does it mean when I get a negative under the square?

When you get something like $ (x+3)^2 = -4 $, it means no real solutions exist. You can't square any real number and get a negative result!

Q: Are there any solutions at all to this equation?

There are complex solutions involving imaginary numbers, but in basic algebra we focus on real solutions only. For real numbers, this equation has no x-intercepts.

Q: How is this different from a parabola that does cross the x-axis?

Compare with $ y = (x+3)^2 - 4 $ (notice the minus). This parabola has its vertex at (-3, -4), so it does cross the x-axis at two points because part of it lies below y = 0.

Quadratic Functions with No Real Intersections

Find the intersection of the function

$y=(x+3)^2+4$

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 with the X-axis

00:04 Substitute Y=0 and solve to find the intersection point

00:08 We want to isolate X

00:18 Extract the root

00:23 Any number squared is always equal to a positive number

00:28 Therefore there is no solution to the question

00:46 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

$y=(x+3)^2+4$

With the X

Step-by-step solution

To determine where the parabola intersects the x-axis, we solve for x when $y = 0$ in the function $y = (x+3)^2 + 4$ .

Set $y = 0$ :

$0 = (x+3)^2 + 4$

Subtract 4 from both sides to isolate the square term:

$(x+3)^2 = -4$

We now assess the equation $(x+3)^2 = -4$ . A square is always non-negative, meaning that no real number squared gives a negative result. Hence, there's no real value of x satisfying this equation.

Thus, the parabola has no intersection points with the x-axis.

Therefore, the correct answer is that there is no intersection.

Final Answer

There is no intersection

Key Points to Remember

Essential concepts to master this topic

Rule: Set y = 0 to find x-axis intersections
Technique: Solve (x+3)² = -4 shows impossible negative square
Check: Square of any real number is always ≥ 0 ✓

Common Mistakes

Avoid these frequent errors

Assuming all parabolas must cross the x-axis
Don't automatically expect x-intercepts for every quadratic = wrong assumption! Some parabolas lie entirely above or below the x-axis when the discriminant is negative. Always check if the equation has real solutions by examining if squares equal negative numbers.

Practice Quiz

Test your knowledge with interactive questions

Which equation represents the function:

$$ y=x^2 $$

moved 2 spaces to the right

and 5 spaces upwards.

FAQ

Everything you need to know about this question

Why doesn't this parabola touch the x-axis?

This parabola opens upward with its vertex at (-3, 4). Since the lowest point is 4 units above the x-axis, the entire parabola stays above y = 0 and never intersects it.

How can I tell without solving if there are x-intercepts?

Look at the vertex form $y = (x+3)^2 + 4$ . The "+4" tells you the vertex is 4 units above the x-axis. Since parabolas with positive leading coefficients open upward, this one never goes below y = 4.

What does it mean when I get a negative under the square?

When you get something like $(x+3)^2 = -4$ , it means no real solutions exist. You can't square any real number and get a negative result!

Are there any solutions at all to this equation?

There are complex solutions involving imaginary numbers, but in basic algebra we focus on real solutions only. For real numbers, this equation has no x-intercepts.

How is this different from a parabola that does cross the x-axis?

Compare with $y = (x+3)^2 - 4$ (notice the minus). This parabola has its vertex at (-3, -4), so it does cross the x-axis at two points because part of it lies below y = 0.

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