I have a math test today and I am having a terrible time trying to figure out my math, please someone help me.

Question

rewrite each quadratic function in the vertex graphing form
y=a(x-p)squared+q

here is the quadratic equation
y=x squared-4x-5

also please determine the coordinates of the vertex
the equation of the axis of symmetry
the concavity of the function
the y-intercept
the x-intercept(s) or zeros
the maximum(minimum)value of the function
the domain of the function
the range of the function
a sketch of the graph

your help is appreciated:)
thank you.

please no snide remarks

Answer 1

OMG
You can't do slope intercept questions and now you're tackling quadratics?

I'm sorry - you are really going to have to get a maths teacher to spend a LOT of time with you on all this. I'll go through it here, but I'm just going to leave wiki pages for you to read for the understanding of the steps.

y = x^2 - 4x - 5

This is a quadratic function (describing the graph of a parabola) in general form.

The other ways it can be written are in factored form
y = (x-5)(x+1) where the zeros (x intercepts) can be found by making y = 0
ie (x-5)(x+1) = 0 and solving for x

Y intercept (just like for the line) make x = 0 if this is possible, and it is for the parabola, and see what value(s) of y pop out.

y = 0^2 - 4(0) - 5
leave you to it ...

Vertex form - which is what you ask for y = a(x-p)^2 + q
can be found by completing the square

You need to turn the y = x^2 -4x -5 into something that can be made into (x-p)^2 + whatever...

y = (x^2 -4x + 4) -9
y = (x-2)^2 -9

This form gives the vertex as (p,q) which you can read off the vertex form.

The axis of symmetry is the vertical line passing through the vertex - we talked about slope in the other question I answered for you. This line would have equations x = p (read it off the vertex form)

Concavity of the function - it is concave up - looks like a cup. Other people would call it convex

Concave (or concave down) looks like a frown

Minimum value is found at the vertex

Domain of the function is the possible values of x
This parabola has a y value for ALL values of x

The Range of the function is all the possible values of y
For this parabola y can be anything above the vertex (p,q)

For the picture of the graph you can refer the the picture below or you can go to a function grapher website and get it to draw a picture for you.

∪

Well - be real. If you can't even do y = Mx + B and you want to now put a quadratic into y = A(x-P)^2 + Q

You want us to teach you completing the square, finding the roots of a quadratic function, range, domain, parabolas etc ... ?

And oh yeah - emily's axis of symmetry is wrong. She's got the right formula for it but substituted c instead of a. Read it off the y = A(x-P)^2 + Q form with formula x = P

Answer 2

OK

y = x^2 - 4x - 5

vertex is -b/2a, f(-b/2a)
a=1 b= -4
-(-4)/2 * 1 = 4/2 = 2
f(2) = 2^2 - (4*2) - 5
= 4 - 8 - 5
= -9
so the vertex is (2, -9)

a = 1 the 1 is > 0
so the graph has a minimum value of -9 that occurs at x = 2

to find the x intercept, make the equation equal to zero
x^2 - 4x - 5 = 0
(x - 5) (x + 1) = 0
x - 5 = 0
x = 5
x + 1 = 0
x = -1
so the x intercepts are 5 and -1

to find the y intercepts solve f(0)
f(0) = x^2 -4x - 5
= 0^2 -(4*0) - 5
= 0 - 0 - 5
= -5
so the y intercept is -5


the domain of a quadratic equation is ALWAYS (-infinity, infinity)
the range depends if you have a minimum or maximum value
here we have a mininum of -9
so the range will be [-9, infinity)

cannot sketch it for you

sorry

good luck

Answer 3

y= x^2-4x-5 is a parabola.
To find the coordinates of the vertex you have to find the min of the function.
dy/dx=2x-4=0 <=> x=2
If x=2 then y=2^2-8-5=-9
So the vertex is the point of coordinates (2,-9)
The axis of symmetry is simply the straight line that is parallel to the y-axis and that passes through the vertex, so the equation is x=2.
To study the concavity of a function you should study the sign of (d^2y)/(dx^2); if the sign is positive then the function is concave (as in your case).
The y-intercept is determined by the value of the funcion for x=0 => y=-5
The zeros are the solutions of the equation x^2-4x-5=0 => x=5 and x=-1.
We have already found the minimum value and there is no maximum.
The domain is the set in which the function is defined: in this case the domain is the whole real set.
The range is [-9,+infinite)
Bye

Answer 4

x is the axis of symmetry....
-b/2a....so --4/2(-5) = 4/-10 = -2/5 is your value of x and axis of symmetry.

then you put that answer back into the equation to find y.

thats all i know how to do!!! sorry! good luck!!