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

2 Answers
Nov 15, 2017

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

Explanation:

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

Nov 16, 2017

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

  • the y-intercept
  • the x-intercept(s)
  • the vertex (turning point)

Explanation:

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.

Completing the square gives:

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]}