What are the important points needed to graph #f(x)= 3x²+x-5#?

Question

Ethan Adkins · Accepted Answer

#x_1=(-1-sqrt61)/6#

#x_2=(-1+sqrt61)/6#

are solutions of #f(x)=0#

#y=-61/12#

is the minimum of the function

See explanations below

#f(x)=3x²+x-5# When you want to study a function, what is really important are particular points of your function: essentially, when your function is equal to 0, or when it reaches a local extremum; those points are called critical points of the function: we can determine them, because they solve : #f'(x)=0# #f'(x)=6x+1# Trivially, #x=-1/6#, and also, around this point, #f'(x)# is both positive and negative, so we can conclude that So : #f(-1/6)=3*(-1/6)²-1/6-5# #=3*1/36-1/6-5# #=1/12-2/12-60/12# #f(-1/6)=-61/12# is the function's minimum. Also, let's determine where #f(x)=0# #3x²+x-5=0# #Delta=b²-4ac# #Delta=1²-4*3*(-5)# #Delta=61# #x=(-b+-sqrtDelta)/(2a)# So : #x_1=(-1-sqrt61)/6# #x_2=(-1+sqrt61)/6# are solutions of #f(x)=0# \0/ This is our response!

Nathan Axel · Answer

undefined 1. Identify the function: f(x) = 3x² + x - 5
2. Determine the vertex using the formula: Vertex = (-b/2a, f(-b/2a))
3. Find the y-intercept by substituting x = 0 into the function.
4. Calculate additional points by choosing x-values and evaluating the function.
5. Plot the points on a graph.
6. Sketch the graph by connecting the points to form a smooth curve.