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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 Rbarnes18:

Ok, thank you AZ:

No problem!

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