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 $y=x^2-5x+6$