Answer
(a) opens up
(b) $(2,-8)$
(c) $x=2$
(d) The $y$-intercept is $4$; the $x$-intercepts are $\dfrac{6-2\sqrt6}{3}$ and $\dfrac{6+2\sqrt3}{6}$;
(e) Refer to the graph below.
Work Step by Step
a) Identify $a,b,c$ from the standard form $f(x)=ax^2+bx+c$:
$a=3$
$b=-12$
$c=4$
Since $a$ is positive, then the parabola opens up.
b) Determine the vertex using the formula $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$
$x=-\dfrac{b}{2a}=-\dfrac{-12}{2(3)}=2$
$y=3(2^2)-12(2)+4=-8$
Therefore the vertedx is at $(2,-8)$.
c) The axis of symmetry is the line $x=-\frac{b}{2a}$.:
Thus, the axis of symmetry is $x=2$.
d) Determine the intercepts:
FInd the $y$-intercept by setting $x=0$ then solving for the value of $y$:
$y=3(0^2)-12(0)+4=4$
Find the $x$-intercept/s by setting $y=0$ then solving the value of $x$ using the quadratic formula:
$3x^2-12x+4=0$ (so $a=3, b=-12, $ and $c=4$)
$x=\dfrac{12\pm\sqrt{(-12)^2-4(3)(4)}}{2(3)}\\
x=\dfrac{12\pm\sqrt{144-48}}{6}\\
x=\dfrac{12\pm\sqrt{96}}{6}\\
x=\dfrac{12\pm\sqrt{16(6)}}{6}\\
x=\dfrac{12\pm 4\sqrt 6}{6}\\
x=\dfrac{6\pm 2\sqrt 6}{3}$
$x_1=\dfrac{6-2\sqrt 6}{3}$
$x_2=\dfrac{6+2\sqrt 6}{3}$
e) Graph the function:
Plot the vertex, $x$-intercepts, and the $y$-intercept, then reflect these points about the axis of symmetry $x=2$. Then, connect the points using a smooth curve.
Refer to the graph below.
