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

ok soo u need help

yes

so they want a piont slope form equation

do yk how to find those

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

y-y₁=m(x-x₁)

so do u remember how to find slope

y=mx+b?

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

oh, ok

so basically u need to find the slope of this graph

then this should be very easy

do you know how to find slope

Ok, i cant remember

slope is y2-y1 and x2-x1

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

(2,-3) (2,2)

mk so use what i just showed u and calculate tht

ok

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do that

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

yep so ur slope is what?

dang moderator go away i aint dooin nun bad

(5,0)

yes

dammn now im stuck

so ik its d or c

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

yes

Thank you

yes u do

no prollem

anytime u need help just hmu or dm me

that's wrong

this is a quadratic equation

how am i wrong

ii learned dis Butter

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

(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

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

so where do i get h and k

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

Ok, thank you

No problem!

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