To** find the y-intercept** of a **quadratic function**, I start by evaluating the **function** when the input, x, is zero. This is because the y-intercept is where the graph of the **function** crosses the y-axis, which happens at ( x=0 ).

When working with a **function** in **standard form,** which is **$f(x) = ax^2 + bx + c$**, I simply substitute zero for x and solve for f(0) to find the y-coordinate of the y-intercept.

The result is the constant term c, indicating that the **y-intercept** of **$ax^2 + bx + c$** is at the point (0, c).

In cases where the **function** is in general form or vertex form, **$f(x) = a(x-h)^2 + k$**, finding the **y-intercept** still requires replacing **x** with **zero.** I then solve for y to find the corresponding point on the **graph.**

The **y-intercept** can offer insight into the **function**‘s graph and is a fundamental aspect of understanding the shape and position of the **parabola** it represents.

Stay tuned to discover how this process provides a gateway to a deeper comprehension of **quadratic functions** and their behaviors.

## Determining the Y-Intercept of a Quadratic Equation

When I work with a **quadratic function**, I’m looking at an **equation** that can be represented graphically as a **parabola**.

This is a **U-shaped** curve that either opens **upwards** or **downwards** on a graph. A **quadratic function** typically is in **standard form**, which is $y = ax^2 + bx + c$. One key point on this **graph** is where the parabola crosses the vertical **y-axis**. This point is known as the **y-intercept**.

The **y-intercept** is the value of ( y ) when ( x = 0 ). Since it’s the point where the graph **intersects** the **y-axis,** I also know that the **value of x** is zero at the **y-intercept.** To find the **y-intercept** of the quadratic equation, I substitute 0 for ( x ) into the equation and solve for ( y ).

Here is the step-by-step process:

- Start with the
**standard form**of the**quadratic equation**: $y = ax^2 + bx + c$. - Plug in 0 for ( x ): $y = a(0)^2 + b(0) + c$.
- Simplify the equation: ( y = c ).

This tells me that the **y-intercept** is simply the constant term ( c ) in the **quadratic equation**.

To see this in action, let’s say I have a **quadratic equation** **$y = 3x^2 – 6x + 9$**. When I put in 0 for ( x ), the **equation** becomes $y = 3(0)^2 – 6(0) + 9$, which simplifies to ( y = 9 ). This means the **y-intercept** is the **point** (0, 9).

Step | Process | Result |
---|---|---|

1 | Substitute 0 for ( x ) | ( y = c ) |

2 | Simplify the equation | ( y = 9 ) |

3 | Write the y-intercept as a point | (0, 9) |

Therefore, whenever I need to determine the **y-intercept** for any **quadratic function**, I find it straightforward to just look at the constant term. This method ensures a quick and accurate solution.

## Applications and Practice

When I work with **quadratic functions**, I need to understand the concept of a **y-intercept**. The **y-intercept** is the point where the graph of the **quadratic function** crosses the y-axis.

This is found by setting the x-value to zero in the **function** and solving for y. The standard form of a **quadratic function** is $ y = ax^2 + bx + c $, where the **y-intercept** is the value of ( c ).

To put this into practice, suppose we have a **quadratic function** $f(x) = 2x^2 + 3x + 1$. To find the **y-intercept**, I evaluate the **function** when ( x = 0 ).

x | f(x) |
---|---|

0 | $f(0) = 2(0)^2 + 3(0) + 1$ |

So, the **y-intercept** is ( f(0) = 1 ), which is the point (0, 1) on the graph.

For real-world applications, think about throwing a rock or shooting a basketball. The path it follows is a **parabola,** and the initial position of the ball or rock is the **y-intercept**.

Now, let’s try some **practice problems** together:

- Find the
**y-intercept**of the**quadratic function****$ g(x) = -x^2 + 4x – 4 $**. - A ball is thrown, and its height in meters is modeled by
**$h(t) = -5t^2 + 20t + 2$**, where t is the time in seconds. What is the initial height of the**ball?**

For each problem, I’ll apply the same **method**: plug in ( x = 0 ) or ( t = 0 ) for the respective **functions,** and solve for y or h. By regularly solving these problems, I strengthen my understanding of **quadratic functions** and their **application** in various situations.

## Conclusion

In this guide, I’ve walked you through finding the **y-intercept** of a **quadratic function**. Just to recap, you can determine the **y-intercept** by setting ( x = 0 ) in the quadratic equation of the form $f(x) = ax^2 + bx + c$.

The calculated value of ( f(0) ) is your **y-intercept**, which is the point where the parabola crosses the **y-axis**.

This method always works because the nature of a parabola, irrespective of its orientation, guarantees that it will **intersect** the vertical axis at a single point.

This point is significant for understanding the **function’s graph** and can serve as a quick check when sketching a parabola or when simplifying the process of graph plotting.

Remember, the **y-intercept** is practical: it’s often used in various **applications** of **quadratic functions,** from predicting profits in a business model to **calculating** the **trajectory** of an object in **physics.** It’s a foundational aspect when learning about **quadratics,** providing insight into the initial value of **functions** when they start.

I hope you feel more confident now in identifying the **y-intercept** on the graph of a **quadratic function.** With practice, these calculations will become second nature to you.