Find the Quadratic Equation Passing Through Point (5,4)

Q: Why is it (x-5) when the x-coordinate is positive 5?

In vertex form $ y = (x-h)^2 + k $, the h-value is always subtracted from x. So vertex at (5,4) means h=5, giving us (x-5). Think of it as: what value of x makes the expression zero? When x=5, then (x-5)=0.

Q: How do I know this is the vertex and not just a random point?

The parabola in the diagram shows a perfect U-shape with (5,4) at the lowest point. This is the vertex - where the parabola changes direction. Also, the equation $ y = (x-5)^2 + 4 $ has its minimum value of 4 when x=5.

Q: What if the parabola opened downward instead?

For a downward-opening parabola, you'd have $ y = -(x-5)^2 + 4 $ with a negative sign in front. The vertex would still be (5,4), but it would be the maximum point instead of the minimum.

Q: Can I expand this to standard form?

Yes! $ y = (x-5)^2 + 4 $ expands to $ y = x^2 - 10x + 25 + 4 = x^2 - 10x + 29 $. However, vertex form is often more useful because you can immediately see the vertex coordinates.

Q: How do I check if my equation is correct?

Substitute the vertex coordinates into your equation. For $ y = (x-5)^2 + 4 $, when x=5: $ y = (5-5)^2 + 4 = 0 + 4 = 4 $. This gives us point (5,4), which matches the diagram!

Vertex Form with Coordinate Translation

Find the corresponding algebraic representation of the drawing:

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find the appropriate algebraic representation for the function

00:03 The function is a parabola

00:16 The minimum point is 5 units to the right according to the graph, that's how we find P

00:46 The minimum point is 4 units up according to the graph, that's how we find K

00:57 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Find the corresponding algebraic representation of the drawing:

Step-by-step solution

To solve this problem, follow these steps:

Identify the point related to the parabola, which is given as $(5, 4)$ .
This point is likely the vertex of the parabola. The vertex form equation is $y = (x-h)^2 + k$ .
Substitute the vertex coordinates $(h, k) = (5, 4)$ into the vertex form.

Using these steps, substitute $h = 5$ and $k = 4$ into the vertex form:

$y = (x - 5)^2 + 4$

This matches the given point and reflects the parabola intersecting or having its vertex at (5, 4).

Therefore, the algebraic representation of the drawing is $y = (x-5)^2 + 4$ .

Final Answer

$y=(x-5)^2+4$

Key Points to Remember

Essential concepts to master this topic

Vertex Form: Use $y = (x-h)^2 + k$ for parabolas with vertex at (h,k)
Technique: Substitute vertex (5,4) to get $y = (x-5)^2 + 4$
Check: When x=5, y=(5-5)²+4=0+4=4, confirming point (5,4) ✓

Common Mistakes

Avoid these frequent errors

Confusing vertex coordinates with equation signs
Don't write y = (x+5)² when vertex is at (5,4) = wrong parabola position! The h-value gets subtracted in vertex form, so positive 5 becomes (x-5). Always remember: vertex (h,k) gives y = (x-h)² + k.

Practice Quiz

Test your knowledge with interactive questions

Which equation represents the function:

$$ y=x^2 $$

moved 2 spaces to the right

and 5 spaces upwards.

FAQ

Everything you need to know about this question

Why is it (x-5) when the x-coordinate is positive 5?

In vertex form $y = (x-h)^2 + k$ , the h-value is always subtracted from x. So vertex at (5,4) means h=5, giving us (x-5). Think of it as: what value of x makes the expression zero? When x=5, then (x-5)=0.

How do I know this is the vertex and not just a random point?

The parabola in the diagram shows a perfect U-shape with (5,4) at the lowest point. This is the vertex - where the parabola changes direction. Also, the equation $y = (x-5)^2 + 4$ has its minimum value of 4 when x=5.

What if the parabola opened downward instead?

For a downward-opening parabola, you'd have $y = -(x-5)^2 + 4$ with a negative sign in front. The vertex would still be (5,4), but it would be the maximum point instead of the minimum.

Can I expand this to standard form?

Yes! $y = (x-5)^2 + 4$ expands to $y = x^2 - 10x + 25 + 4 = x^2 - 10x + 29$ . However, vertex form is often more useful because you can immediately see the vertex coordinates.

How do I check if my equation is correct?

Substitute the vertex coordinates into your equation. For $y = (x-5)^2 + 4$ , when x=5: $y = (5-5)^2 + 4 = 0 + 4 = 4$ . This gives us point (5,4), which matches the diagram!

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