Functions

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rachelmaddie

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Can someone please check my work? And I found this graph but I don’t know if it’s correct?
860222B2-BB13-4FCA-B608-BEA260D3355E.pngA89B5450-0065-40C8-8799-367B0FFAE32A.jpeg
Rewrite this function in the vortex form.
f(x) = x^2 + 2x - 3
f(x) = x^2 + 2x + 1 - 1 - 3
f(x) = (x + 1)^2 - 4
The vortex is (-1, -4).
(1)

Comparing (1) to the standard vertex form
f(x) = (x - h)^2 + k
The vertex of function is at (h, k) = (-1, -4).

To find x-intercepts, put f(x) = 0
0 = (x + 1)^2 - 4
—> (x + 1)^2 = 4

Square root on both sides
x + 1 = +_2
x + 1 = -2 or x + 1 = 2
= x = -3 or x = 1
x intercepts: x = (-3, 0) and (1,0)

For y intercept put x = 0
f(1) = (1)^2 - 4 = 1 - 4 = -3
y intercept: y = (0, -3)

Axis of symmetry: -b/2a
In f(x) = x^2 + 2x - 3
a = 1 and b = 2
Axis of symmetry: -2/2(1) = -1
 
The work is fine too (once you fix the spelling so it's always "vertex", not "vortex") -- except:
  • You can find the y-intercept more easily in the original form; and it's f(0), not f(1), so you didn't write what you were thinking.
  • You didn't have to find the axis of symmetry separately; it's the x-coordinate of the vertex, which you already knew.
The one actual error is that the axis is not a number (-1), but a line, so you need to state its equation: x = -1.
 
Dr.Peterson said:
The work is fine too (once you fix the spelling so it's always "vertex", not "vortex") -- except:
  • You can find the y-intercept more easily in the original form; and it's f(0), not f(1), so you didn't write what you were thinking.
  • You didn't have to find the axis of symmetry separately; it's the x-coordinate of the vertex, which you already knew.
The one actual error is that the axis is not a number (-1), but a line, so you need to state its equation: x = -1.
Click to expand...
In which line do I need to state that?
 
rachelmaddie said:
Equation: -b/2a
In f(x) = x^2 + 2x - 3
a = 1 and b = 2
Equation: -2/2(1) = -1

Like this?
Click to expand...

No, it asks for the axis of symmetry, so you have to tell them what the axis of symmetry is:

Axis of symmetry: x = -b/(2a)​
...​
Axis of symmetry: x = -1​

Again, the axis of symmetry is a line, so to say what it is, you have to give that line's equation. in what you wrote, "-b/2a" is not an equation at all, and -2/2(1) = -1 is not the equation of a line.

When you write anything, think about what it means, and whether it means what you intend to say. Don't just write something because it uses the words you think someone wants you to use.
 
Dr.Peterson said:
No, it asks for the axis of symmetry, so you have to tell them what the axis of symmetry is:

Axis of symmetry: x = -b/(2a)​
...​
Axis of symmetry: x = -1​

Again, the axis of symmetry is a line, so to say what it is, you have to give that line's equation. in what you wrote, "-b/2a" is not an equation at all, and -2/2(1) = -1 is not the equation of a line.

When you write anything, think about what it means, and whether it means what you intend to say. Don't just write something because it uses the words you think someone wants you to use.
Click to expand...
Sorry, I got confused. So what changes do I need to make then?
 
I gave you the answer.

If you're not sure, then write it out again and we can check it.
 
Dr.Peterson said:
I gave you the answer.

If you're not sure, then write it out again and we can check it.
Click to expand...
Rewrite this function in the vertex form.
f(x) = x^2 + 2x - 3
f(x) = x^2 + 2x + 1 - 1 - 3
f(x) = (x + 1)^2 - 4
The vertex is (-1, -4).
(1)

Comparing (1) to the standard vertex form
f(x) = (x - h)^2 + k
The vertex of function is at (h, k) = (-1, -4).

To find x-intercepts, put f(x) = 0
0 = (x + 1)^2 - 4
—> (x + 1)^2 = 4

Square root on both sides
x + 1 = +_2
x + 1 = -2 or x + 1 = 2
= x = -3 or x = 1
x intercepts: x = (-3, 0) and (1,0)
equation x = -1

For y intercept put x = 0
f(1) = (1)^2 - 4 = 1 - 4 = -3
y intercept: y = (0, -3)

Axis of symmetry: -b/2a
In f(x) = x^2 + 2x - 3
a = 1 and b = 2
Axis of symmetry: -2/2(1) = -1
 
Dr. Peterson wrote you the right answer about the axis of symmetry in his first message.
 
rachelmaddie said:
What I did was I added equation x = -1
Click to expand...
rachelmaddie said:
...
x intercepts: x = (-3, 0) and (1,0)
equation x = -1

For y intercept put x = 0
f(1) = (1)^2 - 4 = 1 - 4 = -3
y intercept: y = (0, -3)

Axis of symmetry: -b/2a
In f(x) = x^2 + 2x - 3
a = 1 and b = 2
Axis of symmetry: -2/2(1) = -1
Click to expand...

But "equation x = -1" has nothing to do with the x-intercepts! And I told you what that last paragraph should be:

Axis of symmetry: x = -b/2a​
In f(x) = x^2 + 2x - 3​
a = 1 and b = 2​
x = -2/2(1) = -1​
Axis of symmetry: x = -1​
 
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