Find the directrix of the parabola with equation $y=-0.5x^2+2x+2$
$begingroup$
Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$
I did this:
$$a=-0.5, b = 2, c = 2$$
Formula for the directrix is:
$$y=-1/(4a)$$
$$y=-1/(4cdot(-0.5))=3.5$$
This is not right:
What went wrong? What is the proper way to do it?
algebra-precalculus analytic-geometry conic-sections
edited Jan 8 at 12:10
Blue
47.9k870153
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
$endgroup$
add a comment |
$begingroup$
Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$
I did this:
$$a=-0.5, b = 2, c = 2$$
Formula for the directrix is:
$$y=-1/(4a)$$
$$y=-1/(4cdot(-0.5))=3.5$$
This is not right:
What went wrong? What is the proper way to do it?
algebra-precalculus analytic-geometry conic-sections
edited Jan 8 at 12:10
Blue
47.9k870153
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
$endgroup$
1
$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22
$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23
$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28
$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33
add a comment |
$begingroup$
Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$
I did this:
$$a=-0.5, b = 2, c = 2$$
Formula for the directrix is:
$$y=-1/(4a)$$
$$y=-1/(4cdot(-0.5))=3.5$$
This is not right:
What went wrong? What is the proper way to do it?
algebra-precalculus analytic-geometry conic-sections
edited Jan 8 at 12:10
Blue
47.9k870153
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
$endgroup$
Find the directrix of the parabola with equation
$$y=-0.5x^2+2x+2$$
I did this:
$$a=-0.5, b = 2, c = 2$$
Formula for the directrix is:
$$y=-1/(4a)$$
$$y=-1/(4cdot(-0.5))=3.5$$
This is not right:
What went wrong? What is the proper way to do it?
algebra-precalculus analytic-geometry conic-sections
algebra-precalculus analytic-geometry conic-sections
edited Jan 8 at 12:10
Blue
47.9k870153
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
edited Jan 8 at 12:10
Blue
47.9k870153
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
edited Jan 8 at 12:10
Blue
47.9k870153
edited Jan 8 at 12:10
Blue
47.9k870153
edited Jan 8 at 12:10
Blue
47.9k870153
47.9k870153
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
asked Jan 8 at 11:19
Ryan CameronRyan Cameron
697
697
1
$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22
$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23
$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28
$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33
add a comment |
1
$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22
$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23
$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28
$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33
1
1
$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22
$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22
$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23
$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23
$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28
$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28
$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33
$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33
add a comment |
1 Answer
1
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oldest
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$begingroup$
Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
or $$y=4+frac{1}{2}$$
answered Jan 9 at 10:32
user376343user376343
3,3833826
$endgroup$
add a comment |
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$begingroup$
Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
or $$y=4+frac{1}{2}$$
answered Jan 9 at 10:32
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
or $$y=4+frac{1}{2}$$
answered Jan 9 at 10:32
user376343user376343
3,3833826
$endgroup$
add a comment |
$begingroup$
Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
or $$y=4+frac{1}{2}$$
answered Jan 9 at 10:32
user376343user376343
3,3833826
$endgroup$
Complete the square $$begin{aligned} y&=-0.5x^2+2x+2\ &=-0.5(x^2-4x)+2\ &=-0.5(x^2-4x+4-4)+2\ &=-0.5[(x-2)^2-4]+2\&=-0.5(x-2)^2+4end{aligned}$$
or equivalently $$y-4=-0.5(x-2)^2.$$
The directrix is given by $$y-4=-frac{1}{4a}quad text{with}; a=-0.5$$
or $$y=4+frac{1}{2}$$
answered Jan 9 at 10:32
user376343user376343
3,3833826
answered Jan 9 at 10:32
user376343user376343
3,3833826
answered Jan 9 at 10:32
user376343user376343
3,3833826
answered Jan 9 at 10:32
user376343user376343
3,3833826
3,3833826
add a comment |
add a comment |
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1
$begingroup$
The right formula is $y=-1/(4a)+y_0$, where $y_0$ is the ordinate of the vertex.
$endgroup$
– Aretino
Jan 8 at 11:22
$begingroup$
In addition, you miscomputed $-1/(4a)$.
$endgroup$
– Aretino
Jan 8 at 11:23
$begingroup$
@Aretino How did you derive that formula? Is that always the one to use if I want to find the directrix of a parabola. By the way it works now when I use the formula you denoted.
$endgroup$
– Ryan Cameron
Jan 8 at 11:28
$begingroup$
See here: en.wikipedia.org/wiki/Parabola#As_a_graph_of_a_function
$endgroup$
– Aretino
Jan 8 at 11:33