Parabola-GRE doubt
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How can one interpret just by seeing the equation of parabola that it opens UP/DOWN and or centered around $x$- or $y$-axis.
I know parabola equation as $ax^2+bx+c$, I know that it opens up when $a>0$ and down when $a<0$, but I am not sure to about the orientation of parabola over $x$- or $y$-axis.
For example - how could I have interpreted that $y =x^2-2$ will open up and is oriented on $y$-axis.
My biggest doubt is how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
Thanks in advance.
algebra-precalculus analytic-geometry gre-exam
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
asked Aug 19 at 11:00
ishan sharma
11
add a comment |Â
up vote
-2
down vote
favorite
How can one interpret just by seeing the equation of parabola that it opens UP/DOWN and or centered around $x$- or $y$-axis.
I know parabola equation as $ax^2+bx+c$, I know that it opens up when $a>0$ and down when $a<0$, but I am not sure to about the orientation of parabola over $x$- or $y$-axis.
For example - how could I have interpreted that $y =x^2-2$ will open up and is oriented on $y$-axis.
My biggest doubt is how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
Thanks in advance.
algebra-precalculus analytic-geometry gre-exam
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
asked Aug 19 at 11:00
ishan sharma
11
For equations $y=ax^2+bx+c=0$ , the axis of parabola is parallel to $Y$ axis... Vice versa for the other form of equation. And for negative values of $a$ , the parabola would have a maximum value at a particular point...
– Entrepreneur
Aug 19 at 11:02
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Are you asking about determining the axis of symmetry of a parabola?
– N. F. Taussig
Aug 19 at 11:17
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
How can one interpret just by seeing the equation of parabola that it opens UP/DOWN and or centered around $x$- or $y$-axis.
I know parabola equation as $ax^2+bx+c$, I know that it opens up when $a>0$ and down when $a<0$, but I am not sure to about the orientation of parabola over $x$- or $y$-axis.
For example - how could I have interpreted that $y =x^2-2$ will open up and is oriented on $y$-axis.
My biggest doubt is how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
Thanks in advance.
algebra-precalculus analytic-geometry gre-exam
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
asked Aug 19 at 11:00
ishan sharma
11
How can one interpret just by seeing the equation of parabola that it opens UP/DOWN and or centered around $x$- or $y$-axis.
I know parabola equation as $ax^2+bx+c$, I know that it opens up when $a>0$ and down when $a<0$, but I am not sure to about the orientation of parabola over $x$- or $y$-axis.
For example - how could I have interpreted that $y =x^2-2$ will open up and is oriented on $y$-axis.
My biggest doubt is how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
Thanks in advance.
algebra-precalculus analytic-geometry gre-exam
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
asked Aug 19 at 11:00
ishan sharma
11
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
edited Aug 19 at 11:13
N. F. Taussig
38.6k93053
38.6k93053
asked Aug 19 at 11:00
ishan sharma
11
asked Aug 19 at 11:00
ishan sharma
11
asked Aug 19 at 11:00
ishan sharma
11
11
For equations $y=ax^2+bx+c=0$ , the axis of parabola is parallel to $Y$ axis... Vice versa for the other form of equation. And for negative values of $a$ , the parabola would have a maximum value at a particular point...
– Entrepreneur
Aug 19 at 11:02
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Are you asking about determining the axis of symmetry of a parabola?
– N. F. Taussig
Aug 19 at 11:17
add a comment |Â
For equations $y=ax^2+bx+c=0$ , the axis of parabola is parallel to $Y$ axis... Vice versa for the other form of equation. And for negative values of $a$ , the parabola would have a maximum value at a particular point...
– Entrepreneur
Aug 19 at 11:02
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Are you asking about determining the axis of symmetry of a parabola?
– N. F. Taussig
Aug 19 at 11:17
For equations $y=ax^2+bx+c=0$ , the axis of parabola is parallel to $Y$ axis... Vice versa for the other form of equation. And for negative values of $a$ , the parabola would have a maximum value at a particular point...
– Entrepreneur
Aug 19 at 11:02
For equations $y=ax^2+bx+c=0$ , the axis of parabola is parallel to $Y$ axis... Vice versa for the other form of equation. And for negative values of $a$ , the parabola would have a maximum value at a particular point...
– Entrepreneur
Aug 19 at 11:02
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Are you asking about determining the axis of symmetry of a parabola?
– N. F. Taussig
Aug 19 at 11:17
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Are you asking about determining the axis of symmetry of a parabola?
– N. F. Taussig
Aug 19 at 11:17
add a comment |Â
1 Answer
1
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oldest
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how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
The short answer is: it cannot. The $-2$ has nothing to do with finding the axis of the parabola.
If your parabola has an equation of the form $y = ax^2 + bx + c,$
you find the axis of the parabola by looking at the coefficients $a$ and $b.$
Specifically, the axis will be the line $x = -fracb2a,$ because
$$
ax^2 + bx + c =
aleft(x + fracb2aright)^2 + left(c - fracb^24aright).
$$
When the particular parabola you have to deal with is $y = x^2 - 2,$ then in terms of the usual $a,b,c$ form of the equation you have $a = 1,$ $b = 0,$ and $c = -2.$
Therefore $-fracb2a = 0$ and the axis is the line $x = 0,$
that is, the $y$-axis.
Alternatively, you could simply observe that $y = x^2 - 2$ is just the parabola
$y = x^2$ shifted downward $2$ units.
Since $y = x^2$ has $x = 0$ as its axis, so does $y = x^2 - 2.$
(It only went downward, not sideways.)
answered Aug 19 at 16:04
David K
48.8k340109
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
The short answer is: it cannot. The $-2$ has nothing to do with finding the axis of the parabola.
If your parabola has an equation of the form $y = ax^2 + bx + c,$
you find the axis of the parabola by looking at the coefficients $a$ and $b.$
Specifically, the axis will be the line $x = -fracb2a,$ because
$$
ax^2 + bx + c =
aleft(x + fracb2aright)^2 + left(c - fracb^24aright).
$$
When the particular parabola you have to deal with is $y = x^2 - 2,$ then in terms of the usual $a,b,c$ form of the equation you have $a = 1,$ $b = 0,$ and $c = -2.$
Therefore $-fracb2a = 0$ and the axis is the line $x = 0,$
that is, the $y$-axis.
Alternatively, you could simply observe that $y = x^2 - 2$ is just the parabola
$y = x^2$ shifted downward $2$ units.
Since $y = x^2$ has $x = 0$ as its axis, so does $y = x^2 - 2.$
(It only went downward, not sideways.)
answered Aug 19 at 16:04
David K
48.8k340109
add a comment |Â
up vote
0
down vote
how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
The short answer is: it cannot. The $-2$ has nothing to do with finding the axis of the parabola.
If your parabola has an equation of the form $y = ax^2 + bx + c,$
you find the axis of the parabola by looking at the coefficients $a$ and $b.$
Specifically, the axis will be the line $x = -fracb2a,$ because
$$
ax^2 + bx + c =
aleft(x + fracb2aright)^2 + left(c - fracb^24aright).
$$
When the particular parabola you have to deal with is $y = x^2 - 2,$ then in terms of the usual $a,b,c$ form of the equation you have $a = 1,$ $b = 0,$ and $c = -2.$
Therefore $-fracb2a = 0$ and the axis is the line $x = 0,$
that is, the $y$-axis.
Alternatively, you could simply observe that $y = x^2 - 2$ is just the parabola
$y = x^2$ shifted downward $2$ units.
Since $y = x^2$ has $x = 0$ as its axis, so does $y = x^2 - 2.$
(It only went downward, not sideways.)
answered Aug 19 at 16:04
David K
48.8k340109
add a comment |Â
up vote
0
down vote
up vote
0
down vote
how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
The short answer is: it cannot. The $-2$ has nothing to do with finding the axis of the parabola.
If your parabola has an equation of the form $y = ax^2 + bx + c,$
you find the axis of the parabola by looking at the coefficients $a$ and $b.$
Specifically, the axis will be the line $x = -fracb2a,$ because
$$
ax^2 + bx + c =
aleft(x + fracb2aright)^2 + left(c - fracb^24aright).
$$
When the particular parabola you have to deal with is $y = x^2 - 2,$ then in terms of the usual $a,b,c$ form of the equation you have $a = 1,$ $b = 0,$ and $c = -2.$
Therefore $-fracb2a = 0$ and the axis is the line $x = 0,$
that is, the $y$-axis.
Alternatively, you could simply observe that $y = x^2 - 2$ is just the parabola
$y = x^2$ shifted downward $2$ units.
Since $y = x^2$ has $x = 0$ as its axis, so does $y = x^2 - 2.$
(It only went downward, not sideways.)
answered Aug 19 at 16:04
David K
48.8k340109
how can $-2$ in $y=x^2-2$ can help in revealing that the parabola equation is centered around $y$-axis.
The short answer is: it cannot. The $-2$ has nothing to do with finding the axis of the parabola.
If your parabola has an equation of the form $y = ax^2 + bx + c,$
you find the axis of the parabola by looking at the coefficients $a$ and $b.$
Specifically, the axis will be the line $x = -fracb2a,$ because
$$
ax^2 + bx + c =
aleft(x + fracb2aright)^2 + left(c - fracb^24aright).
$$
When the particular parabola you have to deal with is $y = x^2 - 2,$ then in terms of the usual $a,b,c$ form of the equation you have $a = 1,$ $b = 0,$ and $c = -2.$
Therefore $-fracb2a = 0$ and the axis is the line $x = 0,$
that is, the $y$-axis.
Alternatively, you could simply observe that $y = x^2 - 2$ is just the parabola
$y = x^2$ shifted downward $2$ units.
Since $y = x^2$ has $x = 0$ as its axis, so does $y = x^2 - 2.$
(It only went downward, not sideways.)
answered Aug 19 at 16:04
David K
48.8k340109
answered Aug 19 at 16:04
David K
48.8k340109
answered Aug 19 at 16:04
David K
48.8k340109
answered Aug 19 at 16:04
David K
48.8k340109
48.8k340109
add a comment |Â
add a comment |Â
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For equations $y=ax^2+bx+c=0$ , the axis of parabola is parallel to $Y$ axis... Vice versa for the other form of equation. And for negative values of $a$ , the parabola would have a maximum value at a particular point...
– Entrepreneur
Aug 19 at 11:02
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. Are you asking about determining the axis of symmetry of a parabola?
– N. F. Taussig
Aug 19 at 11:17