Solve y-5x²=20: Finding Y-Axis Intersection Points

Q: Why do we set x = 0 to find the y-intercept?

The y-axis is where x = 0 on a coordinate plane. Any point on the y-axis has coordinates (0, y). So to find where a curve crosses the y-axis, we substitute x = 0 into the equation.

Q: What's the difference between y-intercept and x-intercept?

For y-intercept: set x = 0 and solve for y. For x-intercept: set y = 0 and solve for x. Y-intercept gives points like (0, 20), while x-intercept gives points like (-2, 0).

Q: Can a function have more than one y-intercept?

No! A function can only cross the y-axis at one point. This is because each x-value (including x = 0) can only have one y-value in a function.

Q: How do I write the y-intercept as a coordinate point?

Always write it as (0, y-value). Since we're on the y-axis, the x-coordinate is always 0. In this problem, when y = 20, the point is (0, 20).

Y-Intercept Finding with Quadratic Equations

At which points does the following function intersect the Y axis?

$y-5x^2=20$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:08 Let's find where the graph crosses the Y-axis.

00:11 Remember, at the X-axis crossing, Y equals zero.

00:16 Let's plug in X equals zero into our equation and solve step by step.

00:30 This gives us the Y value at the crossing.

00:35 We used X equals zero from the start to find this.

00:41 And that's how we solve this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

At which points does the following function intersect the Y axis?

$y-5x^2=20$

Step-by-step solution

To determine where the function intersects the Y-axis, we will follow these steps:

Step 1: Identify the given equation $y - 5x^2 = 20$ .
Step 2: Set $x = 0$ to find the Y-intercept.
Step 3: Solve for $y$ when $x = 0$ .

Let's apply these steps in detail:

Step 1: The equation provided is:

$y - 5x^2 = 20$

Step 2: Since the function intersects the Y-axis when $x = 0$ , substitute $x = 0$ into the equation:

$y - 5(0)^2 = 20$

This simplifies to:

$y - 0 = 20$

Therefore, $y = 20$ .

Step 3: The point where the function intersects the Y-axis is

$(0, 20)$ .

Therefore, the function intersects the Y-axis at point $(0, 20)$ .

Final Answer

$(0,20)$

Key Points to Remember

Essential concepts to master this topic

Y-Intercept Rule: Set x = 0 to find where curve crosses Y-axis
Substitution Technique: Replace x with 0: y - 5(0)² = 20 becomes y = 20
Verification: Check that (0,20) satisfies original equation: 20 - 5(0)² = 20 ✓

Common Mistakes

Avoid these frequent errors

Confusing x-intercept and y-intercept methods
Don't set y = 0 when finding y-intercept = wrong axis intersection! This finds where the curve crosses the x-axis instead. Always set x = 0 to find y-intercept points.

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 do we set x = 0 to find the y-intercept?

The y-axis is where x = 0 on a coordinate plane. Any point on the y-axis has coordinates (0, y). So to find where a curve crosses the y-axis, we substitute x = 0 into the equation.

What's the difference between y-intercept and x-intercept?

For y-intercept: set x = 0 and solve for y. For x-intercept: set y = 0 and solve for x. Y-intercept gives points like (0, 20), while x-intercept gives points like (-2, 0).

Can a function have more than one y-intercept?

No! A function can only cross the y-axis at one point. This is because each x-value (including x = 0) can only have one y-value in a function.

How do I write the y-intercept as a coordinate point?

Always write it as (0, y-value). Since we're on the y-axis, the x-coordinate is always 0. In this problem, when y = 20, the point is (0, 20).

What if I get a negative y-value?

That's completely normal! A negative y-value just means the curve crosses the y-axis below the x-axis. For example, (0, -5) is still a valid y-intercept.

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