Question

What is the equation of the quadratic function with a vertex at (2,–25) and an x-intercept at (7,0)?

Answer 1

Let start with some basic theory of quadratics

If  f(x)=ax2+bx+c  a,b,c ∈R  a≠0    f(x)=ax2+bx+c  a,b,c ∈R  a≠0  has   2    2   roots   r1    r1  and   r2    r2  than it can take the form f(x)=a(x−r1)(x−r2).f(x)=a(x−r1)(x−r2).

The maximum or minimum value of  f(x)    f(x)   is   M=−(b2–4ac)4a  M=−(b2–4ac)4a

If   a>0    a>0  then  M    M  is the minimum value and if   a<0    a<0   then   M    M  is the maximum value.

  

Lets go now to the question

Since  2    2  and   7    7  are the roots our function will be  f(x)=a(x−2)(x−7),  a∈R  f(x)=a(x−2)(x−7),  a∈R

so   f(x)=a(x2