Question

1. a parabola that opens downward has its vertex at (3, 0) and a y - intercept at (0, - 9). the parabola is congruent to the parabola described by the function ( f(x)=x^{2} )

a) what is the equation of the function?
b) draw a graph of the function using key points.
c) what is the axis of symmetry?
d) what are the values of the parameters a, h, and k?

1. which function matches the graph?

Explanation:

Step1: Identify vertex form parameters

The vertex form of a parabola is $f(x)=a(x-h)^2+k$, where $(h,k)$ is the vertex. Given vertex $(3,0)$, so $h=3$, $k=0$. Congruent to $f(x)=x^2$ means $|a|=1$; opens downward so $a=-1$.

Step2: Find the parabola equation

Substitute $a=-1$, $h=3$, $k=0$ into vertex form:
$f(x)=-1(x-3)^2+0 = -(x-3)^2$
Expand to standard form to verify y-intercept:
$f(x)=-(x^2-6x+9) = -x^2+6x-9$
Check y-intercept (x=0): $f(0)=-0+0-9=-9$, which matches.

Step3: Determine axis of symmetry

Axis of symmetry for $f(x)=a(x-h)^2+k$ is $x=h$.

Step4: List parameter values

Extract $a$, $h$, $k$ from the vertex form.

Answer:

a) The equation of the function is $\boldsymbol{f(x)=-(x-3)^2}$ (or expanded form $\boldsymbol{f(x)=-x^2+6x-9}$)
b) Key points for graphing:

• Vertex: $(3, 0)$
• Y-intercept: $(0, -9)$
• Symmetric point of y-intercept over axis of symmetry: $(6, -9)$
• Additional point: $(1, -4)$, $(5, -4)$

(Plot these points and draw a downward-opening parabola through them.)
c) The axis of symmetry is $\boldsymbol{x=3}$
d) The values are $\boldsymbol{a=-1}$, $\boldsymbol{h=3}$, $\boldsymbol{k=0}$