Question

graph the function $f(x)=x^2 + 4x + 3$.
state the equation of the axis of symmetry of $f(x)$.
25.
graph the function $f(x)=-x^2 - 6x$ on the set of axes below.
state the coordinates of the vertex of the graph.

Explanation:

Step1: Find axis of symmetry for $f(x)=x^2+4x+3$

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

Step2: Find vertex for $f(x)=-x^2-6x$

Vertex $x$-coordinate: $x=-\frac{b}{2a}$, $a=-1, b=-6$:
$x=-\frac{-6}{2(-1)}=-3$

Step3: Find vertex $y$-coordinate

Substitute $x=-3$ into $f(x)=-x^2-6x$:
$f(-3)=-(-3)^2-6(-3)=-9+18=9$

Step4: Identify key points for first graph

Roots of $x^2+4x+3=0$: $(x+1)(x+3)=0$, so $x=-1, x=-3$. Vertex at $x=-2$: $f(-2)=(-2)^2+4(-2)+3=4-8+3=-1$, so vertex $(-2,-1)$.

Step5: Identify key points for second graph

Roots of $-x^2-6x=0$: $-x(x+6)=0$, so $x=0, x=-6$. Vertex at $(-3,9)$.

Answer:

1. Axis of symmetry for $f(x)=x^2+4x+3$: $\boldsymbol{x=-2}$
2. Vertex of $f(x)=-x^2-6x$: $\boldsymbol{(-3, 9)}$

Graphing instructions:

• For $f(x)=x^2+4x+3$: Plot points $(-3,0)$, $(-2,-1)$, $(-1,0)$, then draw an upward-opening parabola through them.
• For $f(x)=-x^2-6x$: Plot points $(-6,0)$, $(-3,9)$, $(0,0)$, then draw a downward-opening parabola through them.