What is the equation of the quadratic graph with a focus of (2, 0) and a directrix of y = −12?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

What is the equation of the quadratic graph with a focus of (2, 0) and a directrix of y = −12?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

|dw:1437691840404:dw| any ideas where those two fellas are in the graph? can you graph them there?
|dw:1437691171977:dw|
right?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yeap we also know that the vertex is equidistant from both meaning, the focus point is away from the vertex, the same amount the directrix is away from the vertex let's call that distance "p" so, that means the vertex is "half-way" between the focus and directrix the parabola opens towards the focus, the focus is above the directrix, in this case the parabola opens upwards then |dw:1437692110933:dw|
the distance from the focus to the vertex is 7, the distance from the directrix to the vertex is 7 the parabola is opening upwards, so the distance "p" is positive the squared variable is the "x" then now that we know where the vertex coordinates are at, and the "p" distance then \(\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) plug in your values, and solve for "y"
wouldn't the focus be on the x-axis, not the y-axis? since it is (2,0) not (0,2)
hmmmm hold the mayo.. .maybe .... I did mess up there, lemme recheck that
|dw:1437692548026:dw|
anyhow, the vertex is half-way, so that makes it a 0,-6 and yes, that makes the "p" distance to 6 only \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)
hmmm wait a sec.. dohh. the x coordinate can't be 0 anyhow hehe one sec
|dw:1437692768286:dw| \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 2}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)
ok, we have the vertex, now what do we do to get the equation for it
just plug in the values in the parabola equation focus form
so it ends up as (x-2)^2 = 4*6(y-(-6))
yeap or \(\bf (x-{\color{brown}{ 2}})^2=4{\color{purple}{ (6)}}(y-{\color{blue}{ (-6)}})\implies (x-2)^2=24(y+6) \\ \quad \\ (x-2)^2=24y+144\implies \cfrac{(x-2)^2-144}{24}=y\)
Is there another way to write that out, because none of the above are answer choices
I guess one could distribute the denominator, and end up with \(\cfrac{(x-2)^2-144}{24}=y\implies \cfrac{(x-2)^2}{24}-\cfrac{\cancel{144}}{\cancel{24}}=y\implies \cfrac{(x-2)^2}{24}-6=y\)
you could also, expand the squared binomial, and simplify it with the 144, and then distribute the denominator
my anwer choices are x^2-4-35; -x^2+4+35; -1/24x^2+1/6x-35/6; and 1/24x^2-1/6-35/6
ok one sec
\(\bf \cfrac{(x-2)^2-144}{24}=y\implies \cfrac{[x^2-4x+4]-144}{24}=y \\ \quad \\ \cfrac{x^2-4x+4-144}{24}=y\implies \cfrac{x^2-4x-140}{24}=y \\ \quad \\ \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y\implies ?\)
what?
any... thing.. that confuses there?
what happens to the second and third fractions
yeah... whatever happens to those? so... simplify them, what do you get for the simplification? or 4/24 and 140/24?
oh, ok. The would simplify to 2/12 and 35/8
sorry 1/6
well... 2/12 = 1/6 and.... 140/24 simplifiefs to 35/6 thus \(\bf \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y \\ \quad \\ \cfrac{1}{24}x^2-\cfrac{1}{6}x-\cfrac{35}{6}=y\)
oh, ok. Thanks for the help
yw

Not the answer you are looking for?

Search for more explanations.

Ask your own question