Showing posts with label Quadratics. Show all posts
Showing posts with label Quadratics. Show all posts
Saturday, May 30, 2015
Sunday, May 24, 2015
Determine Algebraically whether f(x) = 1 + x + x^2 is even, odd, or neither
A function is even if f(x) = f(-x), but for our function
f(-x)=1-x+x^2
so it is not even.
To be odd -f(x)=f(-x), and for our function
-f(x) = -1-x-x^2
and so it is also not odd.
Therefore it is neither.
f(-x)=1-x+x^2
so it is not even.
To be odd -f(x)=f(-x), and for our function
-f(x) = -1-x-x^2
and so it is also not odd.
Therefore it is neither.
Tuesday, August 13, 2013
without graphing, find the vertex and the maximum or minimum value of f(x) f(x) = -3/7 (x-8)^2+7
since f(x) is in this form, vertex is: (8,7) and since -3/7 < 0 the vertex is a maximum
Wednesday, July 24, 2013
Find the quadratic equation whose roots are twice the roots of 4x^2+8x-5=0.
ax^2+bx+c
4x^2+8x-5=0
since a*c=-20
we use 10 and -2 as follows
4x^2+10x-2x-5=0
2x(2x+5)-1(2x+5)=0
(2x-1)(2x+5)=0
so x=1/2 or x=-5/2
and so
(x-1)(x+5)=0
x^2-6x-5=0
:)
4x^2+8x-5=0
since a*c=-20
we use 10 and -2 as follows
4x^2+10x-2x-5=0
2x(2x+5)-1(2x+5)=0
(2x-1)(2x+5)=0
so x=1/2 or x=-5/2
and so
(x-1)(x+5)=0
x^2-6x-5=0
:)
Sunday, July 21, 2013
Derive a quadratic equation using the following points: (1, -4), (2, -5), (3, -10)
y = ax^2+bx+c
(1) (1, -4) -> -4=a+b+c ,
(2) (2, -5) -> -5= 4a+2b+c,
(3) (3, -10) -> -10=9a+3b+c
(2)-(1):
-5+4=4a-a+2b-b+c-c
-1=3a+b (or 3a+b=-1)
(3)-(1):
-6=8a+2b (or 4a+b=-3)
taking the difference of these 2 we get
4a-3a+b-b=-3+1 (or a=-2)
and thus b=5
Now from (1):
-4=-2+5+c (so c=-7)
y=-2x^2+5x+-7
:)
(1) (1, -4) -> -4=a+b+c ,
(2) (2, -5) -> -5= 4a+2b+c,
(3) (3, -10) -> -10=9a+3b+c
(2)-(1):
-5+4=4a-a+2b-b+c-c
-1=3a+b (or 3a+b=-1)
(3)-(1):
-6=8a+2b (or 4a+b=-3)
taking the difference of these 2 we get
4a-3a+b-b=-3+1 (or a=-2)
and thus b=5
Now from (1):
-4=-2+5+c (so c=-7)
y=-2x^2+5x+-7
:)
Sunday, July 14, 2013
Find all integers m for which y^2+my+50 can be factored.
factors of 50:
1, 50
2, 25
5, 10
So m can be
{51, -51, 27, -27, 15, -15}
:)
1, 50
2, 25
5, 10
So m can be
{51, -51, 27, -27, 15, -15}
:)
Sunday, June 30, 2013
Sunday, May 12, 2013
for each of the following find: A) the vertex, B)the y intercept, C) the x intercept of f(x)=(x-1)^2-2
This parabola is in standard form - which is nice:
so our vertex is (1,-2)
to get the y-intercept set x=0, then
to get the x-intercept set f(x)=0, then
so that
x-1=2 or x-1=-2
x=0 or x=-1
x-intercepts: {0, -1}
:)
so our vertex is (1,-2)
to get the y-intercept set x=0, then
to get the x-intercept set f(x)=0, then
so that
x-1=2 or x-1=-2
x=0 or x=-1
x-intercepts: {0, -1}
:)
Thursday, May 9, 2013
what is the number of real solutions in -x^2+9x+7=0
We can use the discriminant to determine the number and type of solutions:
since the discriminant is real and not 0 we know that there are 2 real solutions
:)
Two vehicles leave the same corner at the same time and travel at a right angle to each other. One vehicle travels 3 km an hour faster than the other. If after one hour the vehicles are 15 km apart, find the rate of each vehicle.
divide by 2...
so the slower is 9 km an hour
and the faster is 12 km per hour
for what values of k will x^2+k=0, have two nonreal solutions
If we subtract k from both side we get:

this equation will have two, non-real (imaginary) solutions for any real, positive number.
:)
this equation will have two, non-real (imaginary) solutions for any real, positive number.
:)
Tuesday, May 7, 2013
Finding Axis of symmetry
f(x) = (x − 5)2 + 2
i need a axis of symmetry
Your Answer:
If by f(x) = (x − 5)2 + 2
you mean:

then we can use the fact that this equation is in standard form. This means that

gives us what we need to determine the axis of symmetry.
Therefore the axis of symmetry is:
x=5
:)
you mean:
then we can use the fact that this equation is in standard form. This means that
gives us what we need to determine the axis of symmetry.
Therefore the axis of symmetry is:
x=5
:)
Tuesday, April 23, 2013
Factoring Quadratics expressions,
Factoring quadratics usually gives students headaches, bu it does not have to be this way. Let's go over some facts that will help us.
- Not every quadratic can be factored into binomials (unless you factor over the complex numbers)
- Factoring may be sometimes called finding the zeros, or solving for x
- Special factoring identities are very useful - so try them first
- There are several valid factoring methods but they all depend on finding a pair of factors of the product of the coefficients a and c and that add to the coefficient b
I have been posting several factoring examples, look for them under the labels factoring, factoring polynomials, factoring quadratics, etc.
Thursday, April 18, 2013
Complete the square
Complete the square. Fill in the number that makes the polynomial a perfect-square quadratic.
d2 + 14d +
Solution:
So we add 49 to get a perfect square
d2 + 14d +
Solution:
So we add 49 to get a perfect square
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