(i)Express x^2 - 3x + 5 in the form (x-a)^2 + b
(ii) sketch the graph of y = x^2 - 3x + 5
For (i) i got (x-1.5)^2 + 11/4
I encountered some problems doing the second part(graph sketching). from the above equation, we can determine that the minimum point for the graph is (1.5, 2.75). now sub x = 0 into equation and you get y= 5. but when you sub y=0, you cant get any answer cuz it's a maths error. How do i go about drawin gthe graph since I've only had 2 points?
you cant find the other point cuz the graph doesnt cut the x-axis
so you should just draw the min pt and y-intercept without the curve intercepting the x-axis.
also, note that b^2 - 4ac < 0 , which means it is either always positive or always negative. which in this case, it is always positive.
hope this helps :)
Starmoonsun explained it perfectly.
Add on:
it is always positive because
(i) the coefficient of x^2 is positive
(ii) the equation has been simplified to (x-1.5)^2 + 11/4
Regards,
Eagle
Owner of Strategic Tuition
You only need to find the points which cut the axis. In this case since the point that cuts the x axis is non-existent, you dont have to put it in. Just leave the 2 points.
Originally posted by starmoonsun:you cant find the other point cuz the graph doesnt cut the x-axis
so you should just draw the min pt and y-intercept without the curve intercepting the x-axis.
also, note that b^2 - 4ac < 0 , which means it is either always positive or always negative. which in this case, it is always positive.
hope this helps :)
thanks all! may i know when i will know whether it is positive or negative when b^2 - 4ac < 0? i think i learnt this in A maths before cuz it seems so familiar.
x^2 - 3x + 5
let a =1 (coeff of x^2)
b = -3 (coeff of x)
c = 5
b^2-4ac = (-3)^2 -4(1)(5) = -11 < 0
Originally posted by bonkysleuth:
thanks all! may i know when i will know whether it is positive or negative when b^2 - 4ac < 0? i think i learnt this in A maths before cuz it seems so familiar.
Because the roots of the equation is {-b +/- sqrt(b^2-4ac)} / 2a
if b^2-4ac < 0, you cannot square root it. Hence, there's no real solution for the roots of the equation.
The graph that does not have a real solution for their roots are graphs that do not touch the x-axis. They can be either all positive, or all negative.
Whether it is positive or negative can be seen from the coefficient of x^2. If the coefficient is positive, the graph is positive. If it is negative, the graph will be negative.
Regards,
Eagle
Owner of Strategic Tuition