Answer:
[tex]a=0.5[/tex]
[tex]b=-2[/tex]
Step-by-step explanation:
we have
[tex]y=ax^{2}+bx+3[/tex]
For x=2, y=1
substitute
[tex]1=a(2)^{2}+b(2)+3[/tex]
[tex]4a+2b=-2[/tex] -----> equation A
Remember that
If the function has a relative minimum in (2,1), then the first derivative of the function must be equal to zero when x=2
The first derivative is equal to
[tex]\frac{dy}{dx}=2ax+b[/tex]
so
[tex]0=2a(2)+b[/tex]
[tex]b=-4a[/tex] ----> equation B
Solve the system of equations A and B by substitution
Substitute equation B in equation A
[tex]4a+2(-4a)=-2[/tex]
solve for b
[tex]4a-8a=-2[/tex]
[tex]-4a=-2[/tex]
[tex]a=0.5[/tex]
Find the value of b
[tex]b=-4a[/tex] ----> [tex]b=-4(0.5)=-2[/tex]
The quadratic equation is
[tex]y=0.5x^{2}-2x+3[/tex]
