How do you use the important points to sketch the graph of #y=x^2+4x+6#?

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Answer 1 · 2017-11-15T16:47:19

Put in zero, 1, -1, 2, -2 for x

Putting in x for zero give the value for y where the curve begins.
y = 6

then putting in the small values for x fives a points to start the sketch of the graph

When x = 1 y = 11
When x = -1 y = 3

When x = 2 y = 18
When x = -2 y = 2

Answer 2 · 2017-11-16T18:23:32

The important points which you need for sketching a curve are:

#y=x^2+4x+6# is the equation of a parabola.

To find the #y#-intercept, make #x=0#

#y=(0)^2 +(0) +6" "rarr y =6" "# The point is #(0,6)#

To find the #x#-intercept(s), make #y=0# and solve:

#x^2+4x+6=0# does not factorise.

#x^2 +4x +4 =-6 +4#

#(x+2)^2 = -2#

#x+2 = +-sqrt(-2)#

There is no solution for #x#, so the curve does not cross the #x#-axis.

Find the axis of symmetry: #x = (-b)/(2a)#

#x = (-4)/(2xx1) = -2#

The vertex is on the line #x=-2#, find the #y#-value:

#y= (-2)^2+4(-2) +46= 4-8+6= 2#
The vertex is at the point is #(-2,2)#

There is a point which is a reflection of the point #(0,6)# in the line of symmetry. #(-4,6)#

Plot these #3# points and draw a smooth curve to pass through all of them:

graph{y =x^2+4x+6 [-6.997, 3.003, 1.76, 6.76]}