How do you graph, label the vertex, axis of symmetry and x intercepts of $y = (3 x - 1) (x - 3)$ $y=(3x-1)(x-3)$ ?

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How do you graph, label the vertex, axis of symmetry and x intercepts of $y = (3 x - 1) (x - 3)$ $y=(3x-1)(x-3)$ ?

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Answer 1 · 2017-06-23T05:11:09

graph{(3x-1)(x-3) [-20, 20, -10, 10]}
Vertex = ( $\frac{5}{3}$ $5/3$ , $- \frac{8}{3}$ $-8/3$ )
Axis of symmetry = $x = \frac{5}{3}$ $x=5/3$
X-intercepts = $(\frac{1}{3}, 0) and (3, 0)$ $(1/3,0) and (3,0)$

Explanation:

Multiplying the two brackets we get the quadratic,
$3 x^{2} - 10 x + 3$ $3x^2-10x+3$
Comparing with $a x^{2} + b x + c$ $ax^2+bx+c$ , we get
$a = 3, b = - 10, c = 3$ $a=3 , b=-10, c=3$
and the Discriminant = $b^{2} - 4 a c$ $b^2-4ac$ => $64$ $64$
Co-ordinates of Vertex of parabola are ( $- \frac{b}{2 a}$ $-b/(2a)$ , $- \frac{D}{4 a}$ $-D/(4a)$ )
plug the the values of $a, b, c$ $a,b,c$ to get vertex of parabola.

Axis of symmetry is the x-coordinate of Vertex i.e
$x = \frac{5}{3}$ $x=5/3$

x-intercept of the parabola is basically the roots of equation.
Roots can be obtained by equating the function to $0$ $0$ .
$\Rightarrow (3 x - 1) (x - 3) = 0$ $=>(3x-1)(x-3)=0$
either $3 x - 1 = 0$ $3x-1=0$ or $x - 3 = 0$ $x-3=0$ (By Zero Product Rule)
this gives $x = \frac{1}{3}, 3$ $x=1/3 , 3$
These are the x-intercepts.

Answer 2 · 2017-06-23T05:36:30

X-intercept

$x = \frac{1}{3}$ $x=1/3$
$x = 3$ $x=3$
Vertex $(\frac{5}{3}, - \frac{16}{3})$ $(5/3, -16/3)$
Axis of symmetry $x = \frac{5}{3}$ $x=5/3$

Explanation:

Given -

$y = (3 x - 1) (x - 3)$ $y=(3x-1)(x-3)$

X-intercept
At $y = 0$ $y=0$

$(3 x - 1) (x - 3) = 0$ $(3x-1)(x-3)=0$

$3 x = 1$ $3x=1$
$x = \frac{1}{3}$ $x=1/3$

$x = 3$ $x=3$

$y = 3 x^{2} - x - 9 x + 3$ $y=3x^2-x-9x+3$
$y = 3 x^{2} - 10 x + 3$ $y=3x^2-10x+3$

Vertex

$x = \frac{- b}{2 a} = \frac{- (- 10)}{2 \times 3} = \frac{10}{6} = \frac{5}{3}$ $x=(-b)/(2a)=(-(-10))/(2 xx3)=10/6=5/3$

At $x = \frac{5}{3}$ $x=5/3$

$y = 3 {(\frac{5}{3})}^{2} - 10 (\frac{5}{3}) + 3$ $y=3(5/3)^2-10(5/3)+3$
$y = 3 (\frac{25}{9}) - \frac{50}{3} + 3$ $y=3(25/9)-50/3+3$
$y = \frac{25}{3} - \frac{50}{3} + 3$ $y=25/3-50/3+3$
$y = \frac{25 - 50 + 9}{3} = - \frac{16}{3}$ $y=(25-50+9)/3=-16/3$

Vertex $(\frac{5}{3}, - \frac{16}{3})$ $(5/3, -16/3)$

Axis of symmetry $x = \frac{5}{3}$ $x=5/3$

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How do you graph, label the vertex, axis of symmetry and x intercepts of $y = (3 x - 1) (x - 3)$ $y=(3x-1)(x-3)$ ?

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Explanation:

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How do you graph, label the vertex, axis of symmetry and x intercepts of y=(3x−1)(x−3)y=(3x-1)(x-3)?

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Explanation:

Explanation:

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How do you graph, label the vertex, axis of symmetry and x intercepts of $y = (3 x - 1) (x - 3)$ $y=(3x-1)(x-3)$ ?