What are you trying to do? Are you trying so solve F(x)=0 ? If so, then (x+3)(x-1)=0
I am trying to get to understand this subject but can't seem to grasp it from the books?
I have; y = X^2 + 2x - 3.
This is what I have tried so far;
(x + 2) (x - 1.5)
x = - 2 or x = 1.5
- 2 + 1.5 = - 0.25
2
y = - 0.25^2 + 2 ( -0.25) - 3
( -4, -1)
I know it's wrong but can't see why?
Any help appreciated
Thanks
Do you want to calculate the zero's? ...
What have you done to factorize the quadratic equation? Have you used the quadratic formula? Or? ...
Because:
(x+2)(x-1,5)=x^2+2x-1,5x-3=x^2+0,5x-3 (and this one is not the given equation).
Use:
D=b^2-4.a.c=4-4.(1.(-3))=4+12=16
Can you now calculate the zero's? ...
You can use the calculate the vertex of the parabola:
x=(-b)/2a
You get the x-coordinat of the vertex and you just have to enter the value of the x-coordinate in the equation and you'll find the y-coordinate of the vertex.
Or you can calculate the first derivative and take f'(x)=0.
You have the values, in general a quadratic equation looks like: y=ax^2+bx+c.
In this case the given equation: y=x^2+2x-3
That means: a=1,b=2,c=-3
If you want to use the first derivative, you'll get:
f'(x)=2x+2
and f'(x)=0 so
2x+2=0 <-> 2x=-2 <->x=-1
Now you've the x-coordinate of the vertex, by entering in the equation you'll get the y-coordinate:
y=(-1)^2+2(-1)-3=1-2-3=-4
So the vertex of the parabola: (-1,-4)
Try it with the other method.
You're welcome! Haven't you learned derivatives yet? ...
In that case, do you see now you can just read a,b,c out of the equation and calculate the vertex that way?
Another way is to wright your equation in this form:
y=(x-p)^2+q
Now the coordinates of the top are: (p,q).
In this case:
y=x^2+2x-3
you can write it as: y=(x+1)^2-4
So the coordinates of the vertex are (-1,-4).
It isn't a problem you haven't learned derivatives yet, but they can be very useful in this case. With derivatives you've also the ability (if it's possible offcourse) to calculate the vertex of many other functions.
Post #2 tells you how to find the x-intercepts. For a parabola, you should know that the turning point lies on the axis of symmetry, which is a vertical line halfway between the x-intercepts.
So you should be able to get the x-coordinate easily. Then substitute that value into the given parabola rule to get the y-coordinate.
I'm sure your class notes and textbook would have examples to follow - have you refered to them?
Go read up:
The Quadratic Formula Explained