Determine Point A on the Quadratic Function: f(x) = x² - 8x + 16

Q: How do I know point A is the y-intercept and not another point?

Look at the graph! Point A is positioned on the y-axis where x = 0. The y-intercept is always where the graph crosses the y-axis, which happens when x = 0.

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

The vertex is the lowest (or highest) point of the parabola - here it's at (4,0). The y-intercept is where the graph crosses the y-axis at x = 0, which is point (0,16).

Q: Why do I substitute x = 0 into the function?

Because the y-intercept occurs when x equals zero. By substituting x = 0 into $ f(x) = x^2 - 8x + 16 $, you find the y-coordinate where the graph crosses the y-axis.

Q: Can I factor this function to make it easier?

Yes! $ f(x) = x^2 - 8x + 16 = (x-4)^2 $. This shows the vertex is at (4,0), but for the y-intercept, you still need to evaluate at x = 0: $ (0-4)^2 = 16 $.

Q: How can I double-check my answer?

Substitute your point back: if A = (0,16), then $ f(0) = 0^2 - 8(0) + 16 = 16 $ ✓. Also, visually check that point A is positioned at height 16 on the y-axis in the graph.

Y-intercept Calculation with Quadratic Functions

The following function has been graphed below:

$f(x)=x^2-8x+16$

Calculate point A.

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the coordinates of point A

00:03 Point A is the intersection point with the Y-axis of the function

00:08 At the intersection point with Y-axis, X = 0

00:12 Substitute X=0 in the function and solve to find Y value

00:27 This is the Y value at point C

00:35 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

The following function has been graphed below:

$f(x)=x^2-8x+16$

Calculate point A.

Step-by-step solution

Let's solve the problem by following the outlined analysis:

Step 1: Identify important points on the parabola.
Step 2: Calculate the y-intercept by evaluating $f(0)$ .
Step 3: Confirm understanding of the vertex form and its characteristics.

Step 1: Identify the significant points on the function.
The function given is $f(x) = x^2 - 8x + 16$ .

This function can be seen as
$f(x) = (x - 4)^2$ .

This format not only indicates it is always non-negative but also reveals the vertex is located at $x = 4$ , importantly, with $f(x) = 0$ .

Step 2: Calculate the y-intercept.
Evaluate the function at $x = 0$ :

$f(0) = 0^2 - 8 \cdot 0 + 16 = 16$ .
So, the y-intercept is $(0, 16)$ .

Thus, point A, which is often labeled at a crucial intercept, corresponds to the y-intercept of the function. The calculation confirms that point A is $(0, 16)$ .

Therefore, the solution to the problem is $(0,16)$ .

Final Answer

$(0,16)$

Key Points to Remember

Essential concepts to master this topic

Y-intercept Rule: Set x = 0 and evaluate f(0) to find y-intercept
Substitution: $f(0) = 0^2 - 8(0) + 16 = 16$
Verification: Point (0,16) lies on y-axis where graph crosses ✓

Common Mistakes

Avoid these frequent errors

Confusing y-intercept with vertex location
Don't find the vertex at x = 4 when asked for point A on the y-axis = wrong coordinates! The vertex is where the parabola reaches its minimum, but point A is clearly on the y-axis in the graph. Always identify what specific point the question is asking for by looking at the graph position.

Practice Quiz

Test your knowledge with interactive questions

The following function has been graphed below:

$$ f(x)=-x^2+5x+6 $$

Calculate points A and B.

FAQ

Everything you need to know about this question

How do I know point A is the y-intercept and not another point?

Look at the graph! Point A is positioned on the y-axis where x = 0. The y-intercept is always where the graph crosses the y-axis, which happens when x = 0.

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

The vertex is the lowest (or highest) point of the parabola - here it's at (4,0). The y-intercept is where the graph crosses the y-axis at x = 0, which is point (0,16).

Why do I substitute x = 0 into the function?

Because the y-intercept occurs when x equals zero. By substituting x = 0 into $f(x) = x^2 - 8x + 16$ , you find the y-coordinate where the graph crosses the y-axis.

Can I factor this function to make it easier?

Yes! $f(x) = x^2 - 8x + 16 = (x-4)^2$ . This shows the vertex is at (4,0), but for the y-intercept, you still need to evaluate at x = 0: $(0-4)^2 = 16$ .

How can I double-check my answer?

Substitute your point back: if A = (0,16), then $f(0) = 0^2 - 8(0) + 16 = 16$ ✓. Also, visually check that point A is positioned at height 16 on the y-axis in the graph.

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