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3 years ago

A parabola has a vertex at (4, 0) and passes through the point (6, 1). What is the equation of the parabola?

1 answer:

7 0

We are given the vertex of a parabola equal to (4, 0) and that the passes the point (6,1). In this case, it is assumed from the given that the parabola is facing upwards. In this case, the equation is (y-k) = 4a (x-h)^2 ; y = 4a (x-4)^2 when x is 6 and y is 1. then,
1 = 4a (6-4)^24a is equal to 1/4
the equation then is y = 1/4 (x-4)^2

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Pls only do this if yk it because i’m giving correct answer brainliest! :))

Leno4ka [110]

Answer:

$x=2+\frac{1}{2}\sqrt[]{21}$

$x=2-\frac{1}{2}\sqrt{{21}$

Step-by-step explanation:

$4x^2-16x-26=-21$

Add 21 on both sides.

$4x^2-16x-26+21=-21+21$

$4x^2-16x-5=0$

a=4

b=-16

c=-5

$x=\frac{-b\frac{+}{}\sqrt[]{b^2-4ac} }{2a}$

$x=\frac{-(-16)\frac{+}{}\sqrt[]{(-16)^2-4(4)(-5)} }{2(4)}$

$x=\frac{16\frac{+}{}\sqrt[]{256+80} }{8}$

$x=\frac{16\frac{+}{}\sqrt[]{336} }{8}$

$x=\frac{16\frac{+}{}\sqrt[]{2^2*2^2*21} }{8}$

$x=\frac{16\frac{+}{}2*2\sqrt[]{21} }{8}$

$x=\frac{16\frac{+}{}4\sqrt[]{21} }{8}$

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$x=\frac{16}{8}+\frac{4\sqrt[]{21}}{8}$

$x=2+\frac{1}{2}\sqrt[]{21}$

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$x=\frac{16}{8}-\frac{4\sqrt[]{21}}{8}\\x=2-\frac{1}{2}\sqrt{{21}$

6 0

2 years ago

Read 2 more answers

Evaluate tan(Sin-1(-5/13)). Answer as a fraction please.

dangina [55]

Using pythagorean identities,tan(sin^-1(-5/13))= tan(arcsin(-5/13))

Let A = arcsin(-5/13) = -arcsin(5/13)

Thus, sin A = -5/13 and cos A = sqrt(1 - (5/13)^2) = 12/13.= tan(A)= sin(A) / cos(A)= -5/13 / (12/13)= -5/12

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

3 years ago

Can someone help me please

Oxana [17]

C

3 3/4 = 3.75
21/6 = 3.5

5 0

3 years ago

X/3 -9=-12 what is x

Pavel [41]

The answer for x should be X= -9

7 0

3 years ago