Question
Question 5
Look at the graph of the equation y = x2– 2x – 7.
a graph of the quadratic equation y = (x − h)2 − c
How does this graph compare with the graph of the equation y=(x-1)^2-8
?
Look at the graph of the equation y = x2– 2x – 7.
a graph of the quadratic equation y = (x − h)2 − c
How does this graph compare with the graph of the equation y=(x-1)^2-8
?
Answers
Answered by
GPT-4o mini
To compare the graphs of the two equations \( y = x^2 - 2x - 7 \) and \( y = (x - 1)^2 - 8 \), we can first rewrite the first equation in vertex form and determine the vertex and characteristics of both equations.
### Step 1: Convert \( y = x^2 - 2x - 7 \) to Vertex Form
The vertex form of a quadratic equation is given by \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.
1. Start with \( y = x^2 - 2x - 7 \).
2. To convert it, we can complete the square:
- Take the coefficient of \( x \) (which is \(-2\)), divide it by 2 to get \(-1\), and square it to get \(1\).
- Adjust the equation by adding and subtracting \(1\):
\[
y = (x^2 - 2x + 1) - 1 - 7
\]
- This simplifies to:
\[
y = (x - 1)^2 - 8
\]
### Step 2: Compare with \( y = (x - 1)^2 - 8 \)
Now we see that both equations, \( y = x^2 - 2x - 7 \) and \( y = (x - 1)^2 - 8 \), are actually the same after rewriting the first equation to vertex form.
### Vertex:
- For both functions, the vertex is \( (1, -8) \).
### Characteristics:
- Both graphs open upwards (as the coefficient of the \( (x - h)^2 \) term is positive).
- The vertex is at the same point: \( (1, -8) \).
- The shape of the parabolas is identical since they are effectively the same quadratic equation.
### Conclusion:
The graphs of \( y = x^2 - 2x - 7 \) and \( y = (x - 1)^2 - 8 \) are the same. They overlap perfectly, as they represent the same quadratic function in different forms.
### Step 1: Convert \( y = x^2 - 2x - 7 \) to Vertex Form
The vertex form of a quadratic equation is given by \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.
1. Start with \( y = x^2 - 2x - 7 \).
2. To convert it, we can complete the square:
- Take the coefficient of \( x \) (which is \(-2\)), divide it by 2 to get \(-1\), and square it to get \(1\).
- Adjust the equation by adding and subtracting \(1\):
\[
y = (x^2 - 2x + 1) - 1 - 7
\]
- This simplifies to:
\[
y = (x - 1)^2 - 8
\]
### Step 2: Compare with \( y = (x - 1)^2 - 8 \)
Now we see that both equations, \( y = x^2 - 2x - 7 \) and \( y = (x - 1)^2 - 8 \), are actually the same after rewriting the first equation to vertex form.
### Vertex:
- For both functions, the vertex is \( (1, -8) \).
### Characteristics:
- Both graphs open upwards (as the coefficient of the \( (x - h)^2 \) term is positive).
- The vertex is at the same point: \( (1, -8) \).
- The shape of the parabolas is identical since they are effectively the same quadratic equation.
### Conclusion:
The graphs of \( y = x^2 - 2x - 7 \) and \( y = (x - 1)^2 - 8 \) are the same. They overlap perfectly, as they represent the same quadratic function in different forms.