How do you write a quadratic function in intercept form whose graph has x intercepts 2, 3 and passes through (4,2)?

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Answer 1 · 2017-06-29T13:33:18

Quadratic function is #y=x^2-5x+6#

Asthe #x#-intercepts are #2,3#, the function passes through #(2,0)# and #(3,0)#. It also passes through #(4,2)#.

Now let the quadratic function be #y=ax^2+bx+c#.

As the function passes through points #(2,0), (3,0)# and #(4,2)#, we have

#0=a*2^2+b*2+c# or #4a+2b+c=0# ...........(A)

#0=a*3^2+b*3+c# or #9a+3b+c=0# ...........(B)

and #2=a*4^2+b*4+c# or #16a+4b+c=2# ...........(C)

Subtracting (A) from (B) and (B) from (C) we get following two equations

#5a+b=0# ...........(1) and

#7a+b=2# ...........(2)

Subtracting (1) from (2), we get #2a=2# i.e. #a=1#

and putting it in (1), we get #b=-5#

Now putting values of #a# and #b# in (A), we get #4-10+c=0# or #c=6#

Hence, quadratic function is #y=x^2-5x+6#

graph{x^2-5x+6 [-1.333, 8.667, -1.12, 3.88]}

Alternatively , a shorter method is that as intercepts are #2# and #3#,

the quadratic function in intercept form is #y=a(x-2)(x-3)#

and as it passes through #(4,2)#, we have

#2=a*(4-2)(4-3)# or #2a=2# i.e. #a=1#

and function is #y=x^2-5x+6#