Y 4x X 2 3

looks like your y-intercepts is at (0,-iii).

looks like your ten-intercepts are at (1,0) and (three,0).

looks like your axis of symmetry is at 10 = 2.

looks similar your vertex is at (two,i).

coordinates are shown as (x,y) where x is the value along the ten-axis and y is the value along the y-centrality.

the point (ii,ane) is the intersection of a vertical line at x = 2 and a horizontal line at y = one.

yous find the x-intercepts by solving for the roots of the equation.

your equation is -10^2 + 4x - iii.

to find the roots you set up the equation equal to 0.

you get -ten^2 + 4x - 3 = 0

information technology's easier to factor if the coefficient of the 10^2 term is positive so multiply both sides of the equation by -1 to go:

x^ii - 4x + 3 = 0

multiplying both sides of an equality by -1 keeps the equality so you lot are changing the equation, but not changing the equal relations between the 2 sides of the equation.

the equal human relationship is what's important to solving for the roots, and so changing the equation past multiplying both sides past -i will requite you the same answer.

down below, at the cease, i'll show y'all how the factors worked out if you did not multiply both sides of the eqution by -one before factoring.

your factors appear to be (x-3) * (x-1) = 0

if you multiply (x-three) * (x-1), yous volition get x^2 -x -3x + 3 which simplifies to x^two - 4x + iii.

this makes the value of y equal to 0 then x = 3 or 10 = 1.

plug 3 into your original equation and you will see that information technology volition equal to 0.

plus 1 into your original equation and you will meet that it will likewise equal to 0.

the graph confirms that the values of 10 are one and 3 when the value of y is equal to 0.

since the coefficient of your 10^two term in the original equation is negative, the graph of the equation will indicate up and open down.

that means that the vertex of the equation will be a maximum point.

if the coefficient of your 10^2 term was positive, the graph ofr the equation would have pointed down and open up.

that means that the vertex of the equation would take been a minimum point.

but it's not.

it'due south a maximum indicate considering the coefficient of the x^2 term is negative.

the vertex of your equation is establish past the equation:

x = -b/2a.

once your equation is in standard form, you lot tin can find the value of a, b, and c.

the standard form of a quadratic equation is ax^ii + bx + c = 0

your equation, in standard form, is -ten^2 + 4x - 3 = 0.

that makes a = -1, b = 4, c = -3

a is the coefficient of the x^2 term.
b is the coefficient of the ten term.
c is the coefficient of the abiding term.

the formula for the vertex of a quadratic equation is ten = -b/2a.

in your equation, that translates to ten = - (-iv) / 2 which becomes 4/2 which becomes 2.

the y value of the vertex is the value of y when x = 2.

that would be -x^2 + 4x - iii = -(two^2) + 8 - 3 = -iv + 8 - 3 = 4 - 3 = 1.

that puts the vertex of your quadratic equation at (x,1) as confirmed by the graph.

the axis of symmetry of your graph is the x value of your vertex.

that makes the axis of symmetry of your graph at x = 2.

to prove that it is the axis of symmetry, yous take a value of y and solve for x.

you can choose value of y = 0 to go far easy.

when y = 0, x = 1 and x = 3.

that makes the value of 10 equidistant from the centrality of symmetry as it should exist.

the y-intercept is the value of y when x = 0.

to find that, simply replace 10 with 0 in your equation to become y = -10^two + 4x - 3 becomes y = 0 + 0 - 3 which becomes y = -iii equally confirmed by the graph.

to plot the graph, you simply take values of ten and solve for y and and so plot the value of the (x,y) pairs on the graph.

a elementary table beneath shows y'all what you lot will notice.

          X	Y = -X^2 + 4X - 3  -two	-15 -1	-8 0	-iii i	0 two	1 iii	0 4	-3 v	-viii vi	-15        

Since the axis of symmetry is at x = 2, you tin can see that:

when 10 = 0 and x = 4, the value of y is equal to -3.

that's considering 0 is two units away from 2 and four is also 2 units away from 2.

since the graph is symmetric about the value of x = 2, the value of y at those points equi-distant from the axis of symmetry will exist the aforementioned.

same goes for y = -15 when x = 6 and x = -2.

same goes for y = -8 when 10 = 5 and x = -1.

dorsum to factoring.

if you lot did non multiply both sides of the equation by -ane, you tin can still factor it, but it's a little harder to see the factors.

i'll do information technology here just to show you.

your original equation is -ten^2 + 4x - 3.

since the coefficient of the x^2 term is negative, one of your x values has to exist negative, so you will get factors that expect similar:

(-x + a) * (x + b) = -x^ii + 4x - 3

a and b are the constant terms that you desire to find.

since your constant term has to be -three, then you lot will multiply by 3 * -1, or you will multiply by -3 * 1

let's try -iii * 1.

you will get (-x - 3) * (x + 1)

that results in -x^two - x -3x - 3 which results in -x^2 -4x - 3 which is not what we desire.

permit's attempt (-10 + 3) * (10 - i)

that results in -x^ii - x + 3x - 3 which results in -x^2 + 2x - 3 which is not what we want.

permit's try (-10 + one) * (x - three)

that results in -x^2 + 3x + x - iii which results in -x^2 + 4x - iii which is what we want and so we can stop here.

our factors become:

(-ten+one) * (x-iii) = 0

from the factor of (ten-3) = 0, we derive x = 3.

frm the factor of (-x+1) = 0, we derive -x = -1 which results in x = 3 when we multiply both sides of the equation by -1.

whether we multiplied both sides of the quadratic equation in standard class by -one in the start or non, we withal wind upward with the same factors.

it'due south just easier to encounter what the factors are by doing that.

you are left to do that or not at your discretion.

i always found it easier to solve the quadratic equation if the coefficient of the 10^2 term was positive.

Y 4x X 2 3,

Source: https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.346043.html

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