Question

How do you graph `y = x^2 + 4x + 1`?

Answer 1

Find the intercepts and the vertex.

Explanation:

Set x=0 to find the y intercept:

`y = x^2 + 4x + 1`

`y = 0^2 + 0x + 1`

`y=1`

Set y=0 to find the x intercept(s) if they exist:

`y = x^2 + 4x + 1`

`0 = x^2 + 4x + 1`

use the quadratic formula:

`x = (-b+-sqrt(b^2 -4ac))/(2a)`

`x = (-4+-sqrt(4^2 -4*1*1))/(2*1)`

`x=2+-sqrt3`

Now find the Axis of Symmetry (aos):

`aos = -b/(2a) = (-4)/(2*1) = -2`

Find the vertex:

vertex = (aos, f(aos)) `= (-2, f(-2))`

`f(x) = x^2 + 4x + 1`

`f(-2) = (-2)^2 + 4* -2 + 1 = -3`

vertex `= (-2, -3)`

Finally since a > 0 the parabola open up like a smile so the vertex is a minimum, now graph:

graph{x^2 +4x+1 [-11, 9, -3.64, 6.36]}