Question

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

Answer 1

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

Answer 2

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]}