Find the Quadratic Equation: Converting Graph at Point (8,-2) to Algebraic Form

Vertex Form with Downward Opening Parabolas

Find the corresponding algebraic representation of the drawing:

(8,-2)(8,-2)(8,-2)

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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
00:04 The function is a parabola
00:16 The function is sad (facing down), therefore coefficient A is negative
00:48 The maximum point is 8 units to the right according to the graph, that's how we find P
01:25 The maximum point is 2 units down according to the graph, that's how we find K
01:39 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the corresponding algebraic representation of the drawing:

(8,-2)(8,-2)(8,-2)

2

Step-by-step solution

To solve for the algebraic representation of the parabola from the drawing:

  • Step 1: Identify the vertex of the parabola. The drawing indicates the vertex at (8,2) (8, -2) .
  • Step 2: Write the vertex form of the parabola, y=a(xh)2+k y = a(x - h)^2 + k , using the vertex (8,2) (8, -2) as h=8 h = 8 and k=2 k = -2 .
  • Step 3: Determine the orientation of the parabola. The drawing suggests the parabola opens downward, indicating a negative value for a a . Hence, a<0 a < 0 .
  • Step 4: Substitute the vertex and the orientation into the equation: y=1(x8)22 y = -1(x - 8)^2 - 2 , simplifying to y=(x8)22 y = -(x - 8)^2 - 2 .

Therefore, the algebraic representation of the parabola is y=(x8)22 y = -(x - 8)^2 - 2 .

3

Final Answer

y=(x8)22 y=-(x-8)^2-2

Key Points to Remember

Essential concepts to master this topic
  • Vertex Form: Use y=a(xh)2+k y = a(x - h)^2 + k where (h,k) is vertex
  • Direction Check: Downward opening means a < 0, so use negative coefficient
  • Verification: Substitute vertex point (8,-2): 2=(88)22=2 -2 = -(8-8)^2 - 2 = -2

Common Mistakes

Avoid these frequent errors
  • Using positive coefficient for downward parabolas
    Don't write y=(x8)22 y = (x-8)^2 - 2 for a downward parabola = upward opening curve! This creates the opposite orientation from what's shown. Always use a negative coefficient (a < 0) when the parabola opens downward.

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

How do I know if the parabola opens up or down from the graph?

+

Look at the shape of the curve! If it looks like a smile (∪), it opens upward and uses positive a. If it looks like a frown (∩), it opens downward and uses negative a.

What if I can't see the exact vertex coordinates?

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The vertex is the highest or lowest point of the parabola. Look for where the curve changes direction - that's your vertex point (h,k).

Why is the vertex form better than standard form here?

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Vertex form y=a(xh)2+k y = a(x-h)^2 + k directly shows the vertex (h,k), making it perfect for graph-to-equation problems. You can read the vertex right off the graph!

How do I check if my equation is correct?

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Substitute the vertex coordinates into your equation. The left side should equal the right side. Also, check that your parabola opens in the same direction as shown in the graph.

What does the 'a' value control besides direction?

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The absolute value of 'a' controls how wide or narrow the parabola is. Larger |a| makes it narrower, smaller |a| makes it wider. When |a| = 1, it's the standard width.

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