Quadratic functions?

Question

Find maximum value of f(x) = -5x^2-10x+6

Solve x^2 = 4x by graphing and indicating the roots

solve x^2 + 2x = -2 by graphing and indicating the roots

If you could explain how to do these too that would be great so I would know how to work it out when it comes to the test, I read my book but I just don't get how to do it.

Answer 1

Choose a range, say [-5,5] in steps of 1.
find 5x^2-10x+6
e.g. x=2
f(x)=5(2^2)-10(2)+6=6
(2,6) is a point
plot the points and find out for which (x,y) pair the maximum occurs.
you'll see that x=-1 f(x)=11 (-1,11) gives the maximum.

x^2-4x=0
x(x-4)=0
x=0, x=4 are the roots
You can plot x^2-4x as before.

x^2+2x+2=0
In this case, you'll find that there is no solution.

Answer 2

1) max value will be the vertex. Formula for finding the x value of the vertex is x= -b/2a
x= -(-10)/2(-5)
x= 10/-10
x= -1 (this is actually the line of symmetry)
When they ask for the value, they really want to the y-value/f(x) value. Once you have the x (x= -1), plug it into the original equation...
f(x) = -5x^2 -10x +6
f(-1) = -5(-1)^2 - 10 (-1) +6
f(-1) = -5 (1) +10 +6
f(-1) = -5 + 10 + 6
f(-1) = 11
So the maximum value is 11.

2) Solve by graphing... It's easiest to find the roots first. The roots are the zeroes. To find those, it's easiest to factor.
Start by getting everything on one side.
x^2 = 4x
x^2 - 4x = 0
now factor out an x..
x ( x-4) = 0
solve, so that x= 0 or x= 4.
Now for the graph. First, since 0 and 4 are the zeros, that means that (0,0) and (4,0) are part of the graph.
You can fill in the middle either by doing an x,y table (plug in x=1,2,3) or by finding the vertex. You can use the x = -b/2a to find the x value of the vertex, as with before. You'd get:
x= -(4-)/2(1)=4/2=2
For the y-value, use y= x^2 -4x
y= 2^2-4(2)= 4-8 = -4
so the vertex is (2, -4). That should be enough to graph.
Course, I just now realized (silly me) that they actually want you to find the zeros by graphing, so you'd use the vertex and line of symmetry, and plug in points on either side of it until you get the zeros (places where the parabola crosses the x-axis). I'm leaving the earlier method up there, because it's a good way to check your answer (plus I think it's easier).

3) find the vertex. start by getting everything on one side to get:
x^2 + 2x +2 =0
x= -b/2a = -2/2(1) = -2/2 = -1
then y= x^2 + 2x +2
so y= (-1)^2 +2(-1) +2
=1 -2 +2
=1
vertex is (-1, 1)
(line of symmetry is x= -1)
plug in points on either side of x= -1 to get the rest of the parabola and the zeros.

Answer 3

First Problem the Maximum Value of f(x)= -5x^2 -10x +6
make f(x) =0
Use the formula:
H=-b/2a
a is the value with x^2 in this case 5 a=-5
b the value with the x in this -10
h= -(-10)/2 (-5)
h=10/-10
h= -1
plug it into the equation so
-5 (-1)^2- 10(-1)+6
-5 (1) -10(-1) +6 (two negatives make a positive)
-5+10+6=11
maximum value is -1,11


Problem 2
x^2= 4x
make one side =0 so
x^2-4x=0
since there is no third number the third number is 0
so
x^2-4x+0=0
factor it out
(4x+0)(x-1)
so do each pair of (_) separate
so
4x+0=0
4x=0
divide by 4
x= 0/4 which is 0
x=0
x-1=0
add one to both sides
x=1
plug number in to find the y value
so
y= x-4x when x is 0
y= 0-4(0)
y=0-0
y=0

y= x^2 - 4x when x is 1
y=1^2-4(1)
y=1-4
y=-3

ok now find the maximum value making the equations equal to each other
-b/2a
4/2(1)
4/2=2=x
plug it in
x^2-4x=y
2^2-4(2)=y
4-8=-4
y=-4
(2,-4)
put those three dot on the and draw the shape between dots

Third equation:
x^2+2x=-2
add two to both sides
x^2 + 2x + 2 =0
use the quadratic formula
-2 + or - the square root of 2^2+-4(2) /2
so -2 + or - the square root of 4+-8/2
so -2 + or miss the square root of -4/2 (note you can not take the square root of a negative number)
thus there is no solution

Answer 4

Find maximum value of f(x) = -5x^2-10x+6
Max occurs when x = -b/2a = -(-10)/(2*-5) = -1
So max is -5(-1)^2 -10(-1) +6 = 11

Solve x^2 = 4x by graphing and indicating the roots
Since x^2 is positive this parabola has a minimum and is shaped like a U. The axis of symmetry is x =-b/2a = -4/2 = -2.
The vertex lies on the axis of symmetry and so is at (-2, -4).
Since y = x^2-4x =x(x-4) the roots are 0 and -4. The y-intercept is (0,0). This should give you sufficient info to plot the graph.

y = x^2+2x+2= 0
This is another U-shaped parabola.
x= [-2 +/- sqrt(2^2-4(1)(2))]/2
x = [-2 +/- sqrt(-4)]/2
x = -1 +/- i
So the roots are complex and the graph lies entirely above the x-axis. The axis of symmetry is x = -b/2a = -2/2= -1.
The vertex is at (-1, 1).

Answer 5

Easiest way to find maximum value is use derivatives and critical points: f'(x)=-10x-10, -10x-10=0 => x = -1. Since the x^2 coefficient is negative, the graph is rising (pos derivative) until x=-1, then it's going down, so the max is at the point.

f(1) = -5(1)-10(-1)+6 = -5+10+6 = 11.

Answer 6

i would use the formula (-b/2a,f(-b/2a)