Question

question 6
consider the function $f(x)=x^2 - 6x + 3$ graph $f(x)$
what is the vertex of $f$?
what is the equation of the line of symmetry of $f$?
$f$ has a select an answer of
the $x$-intercept(s) of $f$ is/are
the $y$-intercept of $f$ is
for this question give all rational answers as fractions or integers and all irrational answers rounded to 2 decimal places.
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Explanation:

Step1: Find vertex x-coordinate

For $f(x)=ax^2+bx+c$, $x=-\frac{b}{2a}$. Here $a=1, b=-6$:
$x = -\frac{-6}{2(1)} = 3$

Step2: Find vertex y-coordinate

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

Step3: Find line of symmetry

Line of symmetry is $x=-\frac{b}{2a}$, so:
$x=3$

Step4: Identify extrema type

Since $a=1>0$, parabola opens up, so it has a minimum at vertex.

Step5: Find x-intercepts

Set $f(x)=0$, solve $x^2-6x+3=0$ using quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$:
$x=\frac{6\pm\sqrt{(-6)^2-4(1)(3)}}{2(1)}=\frac{6\pm\sqrt{24}}{2}=\frac{6\pm2\sqrt{6}}{2}=3\pm\sqrt{6}$
$\sqrt{6}\approx2.45$, so $x\approx3+2.45=5.45$ and $x\approx3-2.45=0.55$

Step6: Find y-intercept

Set $x=0$, substitute into $f(x)$:
$f(0)=0^2-6(0)+3=3$

Answer:

• Vertex of $f$: $(3, -6)$
• Equation of the line of symmetry: $x=3$
• $f$ has a minimum of $-6$
• The $x$-intercept(s) of $f$ is/are: $(0.55, 0)$ and $(5.45, 0)$
• The $y$-intercept of $f$ is: $(0, 3)$

(To graph: Plot the vertex $(3,-6)$, y-intercept $(0,3)$, x-intercepts $(0.55,0)$ and $(5.45,0)$, draw a symmetric upward-opening parabola with axis $x=3$.)