Solve y = -27 + 3x²: Finding X-Axis Intersection Points

Q: Why do I set the function equal to zero?

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

Q: What does ± mean in the solution?

The plus-minus symbol (±) means there are two solutions: one positive and one negative. When $ x^2 = 9 $, both 3 and -3 work because $ 3^2 = 9 $ and $ (-3)^2 = 9 $.

Q: How do I know which answer choice is correct?

Look carefully! The question asks for intersection points (plural). Since we found both (3,0) and (-3,0), the answer that includes both points is correct.

Q: Can a parabola have more than two x-intercepts?

No! A parabola (quadratic function) can have at most two x-intercepts. It might have two (like this problem), one, or none at all.

Q: What if I get a negative number under the square root?

If $ x^2 $ equals a negative number, there are no real solutions. This means the parabola doesn't touch the x-axis at all - it's either entirely above or below it.

Quadratic Functions with X-Intercepts

Determine the points of intersection of the function

$y=-27+3x^2$

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 points of the function with the X-axis

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

00:08 Substitute Y=0 in our equation and solve to find the intersection points

00:12 Isolate X

00:25 Extract the root

00:31 Remember when extracting a root there are 2 solutions (positive and negative)

00:39 These are the X values

00:42 Y = 0 as we substituted at the beginning

00:49 And this is the solution to the 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

$y=-27+3x^2$

With the X

Step-by-step solution

To solve for the intersection points of the function $y=-27+3x^2$ with the x-axis, follow these steps:

Step 1: Set the function equal to zero because intersections with the x-axis occur when $y = 0$ .
So we solve $-27 + 3x^2 = 0$ .
Step 2: Simplify and solve the equation.
Start with $3x^2 = 27$ .
Step 3: Divide both sides by 3 to isolate $x^2$ .
$x^2 = 9$ .
Step 4: Solve for $x$ by taking the square root of both sides.
Thus, $x = \pm \sqrt{9}$ , so $x = 3$ or $x = -3$ .

Therefore, the points of intersection are $(3, 0)$ and $(-3, 0)$ .

The correct choices from the answer list are 1: $(3,0)$ and 2: $(-3,0)$ . Therefore, answer choice 4: Answers $a$ and $b$ are correct is correct.

Final Answer

Answers $a$ and $b$ are correct

Key Points to Remember

Essential concepts to master this topic

Rule: Set y = 0 to find x-intercepts
Technique: Solve $-27 + 3x^2 = 0$ gives $x^2 = 9$
Check: Substitute: $-27 + 3(3)^2 = -27 + 27 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting negative square root solution
Don't solve $x^2 = 9$ as just x = 3 = missing half the answer! Square root equations always have positive and negative solutions. Always write $x = ±3$ to get both intercepts.

Practice Quiz

Test your knowledge with interactive questions

Find the ascending area of the function

$$ f(x)=2x^2 $$

FAQ

Everything you need to know about this question

Why do I set the function equal to zero?

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

What does ± mean in the solution?

The plus-minus symbol (±) means there are two solutions: one positive and one negative. When $x^2 = 9$ , both 3 and -3 work because $3^2 = 9$ and $(-3)^2 = 9$ .

How do I know which answer choice is correct?

Look carefully! The question asks for intersection points (plural). Since we found both (3,0) and (-3,0), the answer that includes both points is correct.

Can a parabola have more than two x-intercepts?

No! A parabola (quadratic function) can have at most two x-intercepts. It might have two (like this problem), one, or none at all.

What if I get a negative number under the square root?

If $x^2$ equals a negative number, there are no real solutions. This means the parabola doesn't touch the x-axis at all - it's either entirely above or below it.

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