How do you find the important points to graph $y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ ?

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How do you find the important points to graph $y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ ?

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Answer 1 · 2016-02-08T08:26:15

The important points are x-intercepts $(- 1, 0)$ $(-1, 0)$ and $(3, 0)$ $(3, 0)$ , y-intercept $(0, - 3)$ $(0, -3)$ , and the Vertex $(1, - 4)$ $(1, -4)$

Explanation:

From the given equation $y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ , set $x = 0$ $x=0$ then solve for the y-intercept

$y = x^{2} - 2 x - 3$ $y=x^2-2x-3$

$y = 0^{2} - 2 \cdot 0 - 3$ $y=0^2-2*0-3$

$y = - 3$ $y=-3$

From the given equation $y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ , set $y = 0$ $y=0$ then solve for the x-intercept

$y = x^{2} - 2 x - 3$ $y=x^2-2x-3$

$x^{2} - 2 x - 3 = 0$ $x^2-2x-3=0$

by factoring method

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

$x = 3$ $x=3$ and $x = - 1$ $x=-1$ when $y = 0$ $y=0$

so $(3, 0)$ $(3, 0)$ and $(- 1, 0)$ $(-1, 0)$ are x-intercepts

From the given equation $y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ ,by completing the square, find the vertex

$y = x^{2} - 2 x - 3$ $y=x^2-2x-3$

$y = x^{2} - 2 x + 1 - 1 - 3$ $y=x^2-2x+1-1-3$

$y = {(x - 1)}^{2} - 4$ $y=(x-1)^2-4$

$y - - 4 = {(x - 1)}^{2}$ $y--4=(x-1)^2$

the vertex (h, k)= $(1, - 4)$ $(1, -4)$

graph{y=x^2-2x-3[-20,20, -10, 10]}

have a nice day from the Philippines

Answer 2 · 2016-02-08T09:24:46

Axis of symmetry: $x = 1$ $x=1$
Vertex: $(1, - 4)$ $(1,-4)$
X-intercepts: $(- 1, 0)$ $(-1,0)$ and $(3, 0)$ $(3,0)$

Explanation:

$y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ is a quadratic equation in standard form, $a x + b x + c$ $ax+bx+c$ , where $a = 1, b = - 2, c = - 3$ $a=1, b=-2, c=-3$ . The graph of a quadratic equation is a parabola.

You need the axis of symmetry, the vertex, and the x-intercepts.

Axis of Symmetry
The axis of symmetry is an imaginary line dividing the parabola into two equal halves. The formula for the axis of symmetry is $x = \frac{- b}{2 a}$ $x=(-b)/(2a)$ .

Substitute the given values for $a$ $a$ and $b$ $b$ into the formula for the axis of symmetry.

$x = \frac{- b}{2 a}$ $x=(-b)/(2a)$

$x = (- \frac{- 2}{2 \cdot 1}) =$ $x=(-(-2)/(2*1))=$

$x = \frac{2}{2} =$ $x=2/2=$

$x = 1$ $color(green)(x=1)$

This is the axis of symmetry, and it is also the $x$ $x$ value for the vertex.

Vertex
The vertex is the maximum or minimum point of a parabola. Since $a > 0$ $a>0$ , the vertex is the minimum and the parabola opens upward.

Substitute $1$ $1$ for $x$ $x$ into the quadratic equation and solve for $y$ $y$ .

$y = x^{2} - 2 x - 3$ $y=x^2-2x-3$

$y = 1^{2} - (2 \cdot 1) - 3 =$ $y=1^2-(2*1)-3=$

$y = 1 - 2 - 3$ $y=1-2-3$

$y = - 4$ $color(purple)(y=-4)$

The vertex is $(1, - 4)$ $color(green)((1,color(purple)(-4))$ .

X-Intercepts
The x-intercepts are the values of $x$ $x$ that intersect the y-axis. A parabola has two x-intercepts.

Substitute $0$ $0$ for $y$ $y$ in the quadratic equation.

$0 = x^{2} - 2 x - 3$ $0=x^2-2x-3$

Factor $x^{2} - 2 x - 3$ $x^2-2x-3$

Find two numbers that when added equal $- 2$ $-2$ , and when multiplied equal $- 3$ $-3$ . The numbers $1$ $1$ and $- 3$ $-3$ fit the pattern. Rewrite the equation as its factors.

$(x + 1) (x - 3) = 0$ $color(red)((x+1))color(blue)((x-3))=0$

First solve $(x + 1) = 0$ $color(red)((x+1))=0$

Subtract $1$ $color(red)1$ from both sides.

$x = - 1$ $color(red)(x=-1)$

Next solve $(x - 3) = 0$ $color(blue)((x-3))=0$

Add $3$ $color(blue)(3)$ to both sides.

$x = 3$ $color(blue)(x=3)$

The x-intercepts are $(- 1, 0)$ $(-1,0)$ and $(3, 0)$ $(3,0)$ .

graph{y=x^2-2x-3 [-10, 10, -5, 5]}

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How do you find the important points to graph $y = x^{2} - 2 x - 3$ $y=x^2-2x-3$ ?

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