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Question 346043: Graph f(x)=-x^2+4x-3, labeling the y-intercept, vertex, and axis of symmetry
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 
looks like your y-intercepts is at (0,-3).
looks like your x-intercepts are at (1,0) and (3,0).
looks like your axis of symmetry is at x = 2.
looks like your vertex is at (2,1).
coordinates are shown as (x,y) where x is the value along the x-axis and y is the value along the y-axis.
the point (2,1) is the intersection of a vertical line at x = 2 and a horizontal line at y = 1.
you find the x-intercepts by solving for the roots of the equation.
your equation is -x^2 + 4x - 3.
to find the roots you set the equation equal to 0.
you get -x^2 + 4x - 3 = 0
it's easier to factor if the coefficient of the x^2 term is positive so multiply both sides of the equation by -1 to get:
x^2 - 4x + 3 = 0
multiplying both sides of an equality by -1 keeps the equality so you are changing the equation, but not changing the equal relations between the 2 sides of the equation.
the equal relationship is what's important to solving for the roots, so changing the equation by multiplying both sides by -1 will give you the same answer.
down below, at the end, i'll show you how the factors worked out if you did not multiply both sides of the eqution by -1 before factoring.
your factors appear to be (x-3) * (x-1) = 0
if you multiply (x-3) * (x-1), you will get x^2 -x -3x + 3 which simplifies to x^2 - 4x + 3.
this makes the value of y equal to 0 then x = 3 or x = 1.
plug 3 into your original equation and you will see that it will equal to 0.
plus 1 into your original equation and you will see that it will also equal to 0.
the graph confirms that the values of x are 1 and 3 when the value of y is equal to 0.
since the coefficient of your x^2 term in the original equation is negative, the graph of the equation will point up and open down.
that means that the vertex of the equation will be a maximum point.
if the coefficient of your x^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 have been a minimum point.
but it's not.
it's a maximum point because the coefficient of the x^2 term is negative.
the vertex of your equation is found by the equation:
x = -b/2a.
once your equation is in standard form, you can find the value of a, b, and c.
the standard form of a quadratic equation is ax^2 + bx + c = 0
your equation, in standard form, is -x^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 x term.
c is the coefficient of the constant term.
the formula for the vertex of a quadratic equation is x = -b/2a.
in your equation, that translates to x = - (-4) / 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 - 3 = -(2^2) + 8 - 3 = -4 + 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, you take a value of y and solve for x.
you can choose value of y = 0 to make it easy.
when y = 0, x = 1 and x = 3.
that makes the value of x equidistant from the axis of symmetry as it should be.
the y-intercept is the value of y when x = 0.
to find that, simply replace x with 0 in your equation to get y = -x^2 + 4x - 3 becomes y = 0 + 0 - 3 which becomes y = -3 as confirmed by the graph.
to plot the graph, you simply take values of x and solve for y and then plot the value of the (x,y) pairs on the graph.
a simple table below shows you what you will find.
X Y = -X^2 + 4X - 3
-2 -15
-1 -8
0 -3
1 0
2 1
3 0
4 -3
5 -8
6 -15
Since the axis of symmetry is at x = 2, you can see that:
when x = 0 and x = 4, the value of y is equal to -3.
that's because 0 is 2 units away from 2 and 4 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 be the same.
same goes for y = -15 when x = 6 and x = -2.
same goes for y = -8 when x = 5 and x = -1.
back to factoring.
if you did not multiply both sides of the equation by -1, you can still factor it, only it's a little harder to see the factors.
i'll do it here just to show you.
your original equation is -x^2 + 4x - 3.
since the coefficient of the x^2 term is negative, one of your x values has to be negative, so you will get factors that look like:
(-x + a) * (x + b) = -x^2 + 4x - 3
a and b are the constant terms that you want to find.
since your constant term has to be -3, then you will multiply by 3 * -1, or you will multiply by -3 * 1
let's try -3 * 1.
you will get (-x - 3) * (x + 1)
that results in -x^2 - x -3x - 3 which results in -x^2 -4x - 3 which is not what we want.
let's try (-x + 3) * (x - 1)
that results in -x^2 - x + 3x - 3 which results in -x^2 + 2x - 3 which is not what we want.
let's try (-x + 1) * (x - 3)
that results in -x^2 + 3x + x - 3 which results in -x^2 + 4x - 3 which is what we want so we can stop here.
our factors become:
(-x+1) * (x-3) = 0
from the factor of (x-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 form by -1 in the beginning or not, we still wind up with the same factors.
it's just easier to see 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 x^2 term was positive.
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