Saturday, January 11, 2014

How do I find any x and y intercepts of f(x) = 2x^2 - 4x + 5? I can't figure out the actual equation for each. If someone could do it in detail...

X and Y intercepts of a line or graph or curve means, where the line or curve or graph  cuts or intersects (or touches in extreme case) X axis or Y axis. The number of intercepting points may be one or more or there may not be any intersections(or intercepts) or touching.


If x and y are related  by an equation,y=f(x), then the x interecept is got by putting y=0  and solving for x value. And the y intercept is got by putting x=0 and  solving for y value.


Finding x intercept:


Put y=0 in y=2x^2-4x+5  and solve for x.


2x^2-4x+5=y. The same equation reversed for my convenience please.


2x^2-4x+5= 0,  as y needs to be put to 0


2(x^2-2x)+5=0, rearranged .


2[(x-1)^2 -1}+5=0, rearranged.


2(x-1)^2-2 =-5.


2(x-1)^2=-5+2 .


2(x-1)^2=-3


(x-1)^2=-3/2


x-1= + or - sqrt(-3/2)


x=1+sqrt(-3/2) or


x=1-sqrt(-3/2).


Both  x values are imaginary. So no real X intercepts.


So the given curve do not cut X axis , as the zeros of the expression, 2x^2-4x+5 are imaginary.


Finding Y intercetpt:


Put x=0


y=2*0^2-4*0+5


y=5.


The given curve cuts (or intercepts) axis at  y = 5.


Point to ponder for you:


Your answer (0,5) is the coordinates of the point, where 2x^2-4x+5 cuts Y axis. The Y cordintate alone 5 where the curve cuts the Y axis is the Y intercept. You are very very nearly correct.


Where is the equation?


f(x) = 2x^2-4x+5 is the equatuation, or


y=2x^2-4x+5 is the equation.


Here x is the variable and for  any value of x , you can find the corresponding value of 2x^2-4x+5 and that is the value of y or f(x).


Example : for x=1, 2x^2-4x+5=2*1^2-4*1+5=2-4+5=3.


So y=3, when x=1.


Hope this helps.

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