Question

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

Answer 1

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

Explanation:

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`