Linear Law - Maths

bonkysleuth SG

(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?

starmoonsun SG

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 :)

eagle SG

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

jayh272416 SG

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.

bonkysleuth SG

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.

starmoonsun SG

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

eagle SG

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