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Mathematics
Rbarnes18:

Determine the equation of the graph, and select the correct answer below. (5 points) parabolic function going up from the left through the point approximately negative two and three fourths comma zero and turning at the point negative one comma three and going down through the point zero comma two and through the point approximately three fourths comma zero and continuing towards infinity Courtesy of Texas Instruments Question 3 options: 1) y = −(x + 1)2 + 3 2) y = −(x − 1)2 + 3 3) y = (x + 1)2 + 3 4) y = (x − 1)2 − 3

Rbarnes18:

Shawnte:

ok soo u need help

Rbarnes18:

yes

Shawnte:

so they want a piont slope form equation

Shawnte:

do yk how to find those

Rbarnes18:

no, i just started the lesson and dont understand how to do them.

Shawnte:

y-y₁=m(x-x₁)

Shawnte:

so do u remember how to find slope

Rbarnes18:

y=mx+b?

Shawnte:

no thats slope intercept form, what we are looking for is point slope form which is y-y₁=m(x-x₁)

Rbarnes18:

oh, ok

Shawnte:

so basically u need to find the slope of this graph

Shawnte:

then this should be very easy

Shawnte:

do you know how to find slope

Rbarnes18:

Ok, i cant remember

Shawnte:

slope is y2-y1 and x2-x1

Shawnte:

|dw:1615164515930:dw|

Shawnte:

so do meh a favor and pick two points on your graph

Rbarnes18:

(2,-3) (2,2)

Shawnte:

mk so use what i just showed u and calculate tht

Rbarnes18:

ok

Shawnte:

|dw:1615164771485:dw|

Shawnte:

do that

Rbarnes18:

2-(-3) = 5 2-2 0

Shawnte:

yep so ur slope is what?

Shawnte:

dang moderator go away i aint dooin nun bad

Rbarnes18:

(5,0)

Shawnte:

yes

Shawnte:

dammn now im stuck

Shawnte:

so ik its d or c

Rbarnes18:

Ok, so now i put my numbers in to this form y-y₁=m(x-x₁)

Shawnte:

yes

Rbarnes18:

Thank you

Shawnte:

yes u do

Shawnte:

no prollem

Shawnte:

anytime u need help just hmu or dm me

AZ:

that's wrong

AZ:

this is a quadratic equation

Shawnte:

how am i wrong

Shawnte:

ii learned dis Butter

AZ:

not a linear line so we don't use \( y - y_1 = m(x-x_1)\) we have to use \(y = (x -h)^2 + k\)

AZ:

(h, k) is the vertex of your line and which way does your parabola open? Is it up or down? If it's opening up like U then it's positive if it's opening down, then it'll be \(y = - (x-h)^2 + k\) where (h, k) is your vertex

Rbarnes18:

it opens down so i would have to use y= -(x-h)^2+k

Rbarnes18:

so where do i get h and k

AZ:

(h, k) is the point which is the vertex

1 attachment
Rbarnes18:

Ok, thank you

AZ:

No problem!

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