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Creating a quadratic equation from a condition
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how can I create an equation that satisfies the following: "x-intercepts 1 and -1, y-intercept 3". I understand that the factored form of a quadratic equation offers both x intercepts however, I'm not sure of a form that offers both x intercepts and y intercepts. Here's another one: x-intercept 3, and passing through the point (1, -2).
quadratics quadratic-forms
asked Jan 17 at 23:19
moemoe
82
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add a comment |
$begingroup$
how can I create an equation that satisfies the following: "x-intercepts 1 and -1, y-intercept 3". I understand that the factored form of a quadratic equation offers both x intercepts however, I'm not sure of a form that offers both x intercepts and y intercepts. Here's another one: x-intercept 3, and passing through the point (1, -2).
quadratics quadratic-forms
asked Jan 17 at 23:19
moemoe
82
$endgroup$
add a comment |
$begingroup$
how can I create an equation that satisfies the following: "x-intercepts 1 and -1, y-intercept 3". I understand that the factored form of a quadratic equation offers both x intercepts however, I'm not sure of a form that offers both x intercepts and y intercepts. Here's another one: x-intercept 3, and passing through the point (1, -2).
quadratics quadratic-forms
asked Jan 17 at 23:19
moemoe
82
$endgroup$
how can I create an equation that satisfies the following: "x-intercepts 1 and -1, y-intercept 3". I understand that the factored form of a quadratic equation offers both x intercepts however, I'm not sure of a form that offers both x intercepts and y intercepts. Here's another one: x-intercept 3, and passing through the point (1, -2).
quadratics quadratic-forms
quadratics quadratic-forms
asked Jan 17 at 23:19
moemoe
82
asked Jan 17 at 23:19
moemoe
82
asked Jan 17 at 23:19
moemoe
82
asked Jan 17 at 23:19
moemoe
82
asked Jan 17 at 23:19
moemoe
82
82
add a comment |
add a comment |
1 Answer
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If you have that the x intercepts are at $-1$ and $1$, you must have a quadratic of the form $y=a(x-1)(x+1)$. Now, when $x=0$, $3=(-1)(1)a$, so your quadratic is $y=-3(x+1)(x-1)$.
For your second one; it must be of the form $y=a(x-3)^2$. Plugging in $(1,-2)$ gives $-2=a(4)$ so $y=frac{-1}{2}(x-3)^2$.
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
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$begingroup$
How did you get -3? What does "a" stand for?
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– moe
Jan 17 at 23:37
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
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– cluelessatthis
Jan 17 at 23:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you have that the x intercepts are at $-1$ and $1$, you must have a quadratic of the form $y=a(x-1)(x+1)$. Now, when $x=0$, $3=(-1)(1)a$, so your quadratic is $y=-3(x+1)(x-1)$.
For your second one; it must be of the form $y=a(x-3)^2$. Plugging in $(1,-2)$ gives $-2=a(4)$ so $y=frac{-1}{2}(x-3)^2$.
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
How did you get -3? What does "a" stand for?
$endgroup$
– moe
Jan 17 at 23:37
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
$endgroup$
– cluelessatthis
Jan 17 at 23:41
add a comment |
$begingroup$
If you have that the x intercepts are at $-1$ and $1$, you must have a quadratic of the form $y=a(x-1)(x+1)$. Now, when $x=0$, $3=(-1)(1)a$, so your quadratic is $y=-3(x+1)(x-1)$.
For your second one; it must be of the form $y=a(x-3)^2$. Plugging in $(1,-2)$ gives $-2=a(4)$ so $y=frac{-1}{2}(x-3)^2$.
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
$endgroup$
$begingroup$
How did you get -3? What does "a" stand for?
$endgroup$
– moe
Jan 17 at 23:37
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
$endgroup$
– cluelessatthis
Jan 17 at 23:41
add a comment |
$begingroup$
If you have that the x intercepts are at $-1$ and $1$, you must have a quadratic of the form $y=a(x-1)(x+1)$. Now, when $x=0$, $3=(-1)(1)a$, so your quadratic is $y=-3(x+1)(x-1)$.
For your second one; it must be of the form $y=a(x-3)^2$. Plugging in $(1,-2)$ gives $-2=a(4)$ so $y=frac{-1}{2}(x-3)^2$.
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
$endgroup$
If you have that the x intercepts are at $-1$ and $1$, you must have a quadratic of the form $y=a(x-1)(x+1)$. Now, when $x=0$, $3=(-1)(1)a$, so your quadratic is $y=-3(x+1)(x-1)$.
For your second one; it must be of the form $y=a(x-3)^2$. Plugging in $(1,-2)$ gives $-2=a(4)$ so $y=frac{-1}{2}(x-3)^2$.
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
edited Jan 18 at 1:24
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
answered Jan 17 at 23:28
cluelessatthiscluelessatthis
402313
402313
$begingroup$
How did you get -3? What does "a" stand for?
$endgroup$
– moe
Jan 17 at 23:37
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
$endgroup$
– cluelessatthis
Jan 17 at 23:41
add a comment |
$begingroup$
How did you get -3? What does "a" stand for?
$endgroup$
– moe
Jan 17 at 23:37
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
$endgroup$
– cluelessatthis
Jan 17 at 23:41
$begingroup$
How did you get -3? What does "a" stand for?
$endgroup$
– moe
Jan 17 at 23:37
$begingroup$
How did you get -3? What does "a" stand for?
$endgroup$
– moe
Jan 17 at 23:37
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
$endgroup$
– cluelessatthis
Jan 17 at 23:41
$begingroup$
The most general form of a quadratic equation is $y=a(x-b)(x-c)$. Here, b and c are the x intercepts, a acts as a vertical ‘stretching’ constant. This allows you to manipulate the y intercept by changing how ‘high’ your function reaches. To get $a=-3$ substitute in $x=0$ and $y=3$ (since a y-intercept of 3 has coordinates $(0,3)$. Your b and c are given in the question as $1$ and $-1$, so substitute those in as well. Then solve for a.
$endgroup$
– cluelessatthis
Jan 17 at 23:41
add a comment |
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