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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