Math Question

[Bacon Boy]

Will a quadratic equation always have an axis of symmetry of x = 0?

I checked Google, no results. When I work things out, I usually get a vertex of (0 , y). There is something in for the y, it's just a placeholder right now.

For example, y = -x^2 + 5x - 3 has a vertex of (0, -3), correct?

 

[Kyel]

6

 

[Bacon Boy]

Disregard.

 

[Mino]

Will a quadratic equation always have an axis of symmetry of x = 0?

I checked Google, no results. When I work things out, I usually get a vertex of (0 , y). There is something in for the y, it's just a placeholder right now.

For example, y = -x^2 + 5x - 3 has a vertex of (0, -3), correct?

Put simply, no.

For example, the function f(x)=(x+1)^2 has an axis of symmetry at -1.

 

[Bacon Boy]

Will a quadratic equation always have an axis of symmetry of x = 0?

I checked Google, no results. When I work things out, I usually get a vertex of (0 , y). There is something in for the y, it's just a placeholder right now.

For example, y = -x^2 + 5x - 3 has a vertex of (0, -3), correct?

Put simply, no.

For example, the function f(x)=(x+1)^2 has an axis of symmetry at -1.

I haven't been inputting enough x and y values, then. I did this so long ago. @_@

 

[Bacon Boy]

However, y = -x^2 + 5x - 3 has a vertex of (0, -3) is correct. Right?

f(0) = -(0)^2 + 5(0) - 3
f(0) = 0 + 0 - 3.
f(0) = -3
Not six, unless Kyel was joking.

 

[Mino]

Will a quadratic equation always have an axis of symmetry of x = 0?

I checked Google, no results. When I work things out, I usually get a vertex of (0 , y). There is something in for the y, it's just a placeholder right now.

For example, y = -x^2 + 5x - 3 has a vertex of (0, -3), correct?

Put simply, no.

For example, the function f(x)=(x+1)^2 has an axis of symmetry at -1.

I haven't been inputting enough x and y values, then. I did this so long ago. @_@

Here's the rules for transforming a quadratic equation (i.e. a parabola).

If we start with the equation y = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2
The red portion will change the parabola's magnitude (I think that's the word for it, could be wrong). In this case, it will make it narrower.

 

[Mino]

However, y = -x^2 + 5x - 3 has a vertex of (0, -3) is correct. Right?

f(0) = -(0)^2 + 5(0) - 3
f(0) = 0 + 0 - 3.
f(0) = -3
Not six, unless Kyel was joking.

Yep. No, sorry. That's actually where it crosses x = 0, not the vertex.

What math are you in? I might have just given you help that was completely unnecessary there.

 

[Bacon Boy]

Will a quadratic equation always have an axis of symmetry of x = 0?

I checked Google, no results. When I work things out, I usually get a vertex of (0 , y). There is something in for the y, it's just a placeholder right now.

For example, y = -x^2 + 5x - 3 has a vertex of (0, -3), correct?

Put simply, no.

For example, the function f(x)=(x+1)^2 has an axis of symmetry at -1.

I haven't been inputting enough x and y values, then. I did this so long ago. @_@

Here's the rules for transforming a quadratic equation (i.e. a parabola).

If we start with the equation y = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2
The red portion will change the parabola's magnitude (I think that's the word for it, could be wrong). In this case, it will make it narrower.

I know that stuff, it's a matter of doing this:

y = -x^2 + 5x - 3

x | y
-1| -9
0 | -3
1 | 1
2 | 3
3 | -3

Obviously, here, the vertex would be (2,3) and the axis would be x = 2. But is there a way to save myself the trouble of inputting the lot of numbers in order the find the vertex and a set of points? In the example, I was just inputting numbers until I found one where the points started dropping instead of increasing.

I'm in Algebra 2.

 

[Mino]

Quoting limited to 4 levels deep

I haven't been inputting enough x and y values, then. I did this so long ago. @_@

Here's the rules for transforming a quadratic equation (i.e. a parabola).

If we start with the equation y = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2
The red portion will change the parabola's magnitude (I think that's the word for it, could be wrong). In this case, it will make it narrower.

I know that stuff, it's a matter of doing this:

y = -x^2 + 5x - 3

x | y
-1| -9
0 | -3
1 | 1
2 | 3
3 | -3

Obviously, here, the vertex would be (2,3) and the axis would be x = 2. But is there a way to save myself the trouble of inputting the lot of numbers in order the find the vertex and a set of points? In the example, I was just inputting numbers until I found one where the points started dropping instead of increasing.

I'm in Algebra 2.

Use a graphing calculator, duh.

 

[Bacon Boy]

Quoting limited to 4 levels deep

Here's the rules for transforming a quadratic equation (i.e. a parabola).

If we start with the equation y = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2
The red portion will change the parabola's magnitude (I think that's the word for it, could be wrong). In this case, it will make it narrower.

I know that stuff, it's a matter of doing this:

y = -x^2 + 5x - 3

x | y
-1| -9
0 | -3
1 | 1
2 | 3
3 | -3

Obviously, here, the vertex would be (2,3) and the axis would be x = 2. But is there a way to save myself the trouble of inputting the lot of numbers in order the find the vertex and a set of points? In the example, I was just inputting numbers until I found one where the points started dropping instead of increasing.

I'm in Algebra 2.

Use a graphing calculator, duh.

That would work too.
Thanks.

 

[Mino]

Quoting limited to 4 levels deepy = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2
The red portion will change the parabola's magnitude (I think that's the word for it, could be wrong). In this case, it will make it narrower.

I know that stuff, it's a matter of doing this:

y = -x^2 + 5x - 3

x | y
-1| -9
0 | -3
1 | 1
2 | 3
3 | -3

Obviously, here, the vertex would be (2,3) and the axis would be x = 2. But is there a way to save myself the trouble of inputting the lot of numbers in order the find the vertex and a set of points? In the example, I was just inputting numbers until I found one where the points started dropping instead of increasing.

I'm in Algebra 2.

Use a graphing calculator, duh.

That would work too.
Thanks.

I feel like there has to be an easy way, though.

 

[Princess]

No. Wherever your vertex is at, the x value is your line of symmetry. (I think that's what you're asking )

 

[Bacon Boy]

Quoting limited to 4 levels deepy = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2

Use a graphing calculator, duh.

That would work too.
Thanks.

I feel like there has to be an easy way, though.

>.< How did I miss this.

Vertex = -b/2a

 

[Mino]

Quoting limited to 4 levels deepy = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2

That would work too.
Thanks.

I feel like there has to be an easy way, though.

>.< How did I miss this.

Vertex = -b/2a

So....

-(5)/2*(-1)

Five halves? (un)

Or am I wrong in assuming that that's referring to this:

ax^2+bx+c

 

[Bacon Boy]

Quoting limited to 4 levels deepy = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2

I feel like there has to be an easy way, though.

>.< How did I miss this.

Vertex = -b/2a

So....

-(5)/2*(-1)

Five halves? (un)

Or am I wrong in assuming that that's referring to this:

ax^2+bx+c

You are correct. Apparently, in a quadratic, the easy way to find the vertex is the negative of the b value over 2 times the a value. Then you plug it in order to get the y value.

 

[Mino]

Quoting limited to 4 levels deepy = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2

>.< How did I miss this.

Vertex = -b/2a

So....

-(5)/2*(-1)

Five halves? (un)

Or am I wrong in assuming that that's referring to this:

ax^2+bx+c

You are correct. Apparently, in a quadratic, the easy way to find the vertex is the negative of the b value over 2 times the a value. Then you plug it in order to get the y value.

I got a 3.25 for the y-value, but yes.

 

[Bacon Boy]

Quoting limited to 4 levels deepy = x^2, then:

y = (x+1)^2
The red portion will move the parabola to the left or right, in this case it will move it to the left one.

y = (x^2)+1
The red portion will move the parabola up or down, in this case it will move it up one.

y = -x^2
The red portion will invert the parabola, in this case making it a bottom-opening parabola.

y = 2x^2

So....

-(5)/2*(-1)

Five halves? (un)

Or am I wrong in assuming that that's referring to this:

ax^2+bx+c

You are correct. Apparently, in a quadratic, the easy way to find the vertex is the negative of the b value over 2 times the a value. Then you plug it in order to get the y value.

I got a 3.25 for the y-value, but yes.

Yea, I didn't take into account that the a value = -1, not 1.

 

[Garrett x50 cal]

Math time with Trevor.