Question

on the set of axes below, graph $f(x) = x^2 + 4x + 1$. state the coordinates of the minimum.

Explanation:

Step1: Find vertex x-coordinate

For $f(x)=ax^2+bx+c$, vertex $x=-\frac{b}{2a}$. Here $a=1, b=4$:
$x=-\frac{4}{2(1)}=-2$

Step2: Find vertex y-coordinate

Substitute $x=-2$ into $f(x)$:
$f(-2)=(-2)^2+4(-2)+1=4-8+1=-3$

Step3: Generate table values

Choose x-values around $x=-2$:

• $x=-4$: $f(-4)=(-4)^2+4(-4)+1=16-16+1=1$
• $x=-3$: $f(-3)=(-3)^2+4(-3)+1=9-12+1=-2$
• $x=-2$: $f(-2)=-3$
• $x=-1$: $f(-1)=(-1)^2+4(-1)+1=1-4+1=-2$
• $x=0$: $f(0)=0^2+4(0)+1=1$
• $x=1$: $f(1)=1^2+4(1)+1=1+4+1=6$
• $x=2$: $f(2)=2^2+4(2)+1=4+8+1=13$

Answer:

Table of Values:

$x$	$y$
-3	-2
-2	-3
-1	-2
0	1
1	6
2	13

Coordinates of the Minimum:

$(-2, -3)$

(To graph: plot the table points and draw a smooth upward-opening parabola through them, with the vertex at $(-2, -3)$)