Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

y=x2+4 can someone help me wth this graph equation i need to find the vertex and the intercepts

OpenStudy (anonymous):

do you mean y = x^2 + 4?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so the standard parabola is y = mx+b where b is the y-intercept so in your equation y = x^2 + 4 so the y-intercept = 4, or (0,4)

OpenStudy (anonymous):

and the x-intercepts are (2,0) and (-2,0) by putting y=0

OpenStudy (dumbcow):

y=0^2 +4 y=4 y_intercept is 4 0 = x^2 +4 -4 = x^2 No real x_intercepts vertex: a=1, b=0, c=4 x=-b/2a x=0/2 = 0 y = 0^2+4 = 4 vertex: (0,4)

OpenStudy (anonymous):

oh pellet..looks lyk an off day

OpenStudy (anonymous):

dumbcows right

OpenStudy (anonymous):

thanks everyone

OpenStudy (anonymous):

i ate a dumbcow at mcdonalds once.... with cheese and a bun and extra pickles... it was mighty tasty :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!