Finding Y-Axis Intersection Points: y=2x²+10 Quadratic Function

Q: Why do we set x = 0 to find y-axis intersections?

The y-axis is the vertical line where x = 0 for every point. To find where any function crosses this line, we substitute x = 0 and solve for y.

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

For y-axis intersections, set x = 0 and solve for y. For x-axis intersections, set y = 0 and solve for x. They give completely different points!

Q: Can a quadratic function have more than one y-axis intersection?

No! Every function can cross the y-axis at most once because there's only one y-value when x = 0. However, it can have multiple x-axis intersections.

Q: How do I write the intersection point correctly?

Always write coordinates as (x, y). For y-axis intersections, the x-coordinate is always 0, so the point is $ (0, y\text{-value}) $.

Q: What if I get a negative y-value?

That's perfectly normal! The y-axis intersection can be above the x-axis (positive y) or below the x-axis (negative y). Just substitute x = 0 accurately.

Y-Axis Intersections with Quadratic Substitution

At which points does the function $y=2x^2+10$

intersect the y axis?

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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 Y-axis

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

00:07 Substitute X=0 in our equation and solve to find the intersection point

00:12 Let's solve and calculate to find the intersection point

00:24 This is the value at the intersection point with the Y-axis

00:28 X = 0 as we substituted at the beginning

00:34 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

At which points does the function $y=2x^2+10$

intersect the y axis?

Step-by-step solution

To solve this problem, we need to determine where the given function intersects the y-axis. This occurs when the x-coordinate is 0. Let's find the y-coordinate when $x = 0$ .

The function in question is $y = 2x^2 + 10$ .
Substitute $x = 0$ into the function:

$y = 2(0)^2 + 10 = 2 \cdot 0 + 10 = 10$ .

Since substituting $x = 0$ yields $y = 10$ , the point of intersection on the y-axis is $(0, 10)$ .

Therefore, the point where the function intersects the y-axis is $(0, 10)$ .

Final Answer

$(0,10)$

Key Points to Remember

Essential concepts to master this topic

Intersection Rule: Y-axis intersections occur when x equals zero
Technique: Substitute x = 0 into $y = 2(0)^2 + 10 = 10$
Check: Point (0,10) has x-coordinate of 0, confirming y-axis intersection ✓

Common Mistakes

Avoid these frequent errors

Confusing x-axis and y-axis intersection methods
Don't set y = 0 when finding y-axis intersections = wrong coordinate system! This gives you x-axis intersections instead. Always set x = 0 for y-axis intersections and substitute into the function.

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 we set x = 0 to find y-axis intersections?

The y-axis is the vertical line where x = 0 for every point. To find where any function crosses this line, we substitute x = 0 and solve for y.

What's the difference between y-axis and x-axis intersections?

For y-axis intersections, set x = 0 and solve for y. For x-axis intersections, set y = 0 and solve for x. They give completely different points!

Can a quadratic function have more than one y-axis intersection?

No! Every function can cross the y-axis at most once because there's only one y-value when x = 0. However, it can have multiple x-axis intersections.

How do I write the intersection point correctly?

Always write coordinates as (x, y). For y-axis intersections, the x-coordinate is always 0, so the point is $(0, y\text{-value})$ .

What if I get a negative y-value?

That's perfectly normal! The y-axis intersection can be above the x-axis (positive y) or below the x-axis (negative y). Just substitute x = 0 accurately.

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