Find the vertex form of a quadratic that passes through (2,6), has an x-int of 1 and a y-int of 6?

The vertex form of a quadratic equation of the form #color(red)(ax^2+bx+c=0)# is given as:

#color(red)(y=a(x-h)^2+k)#

Where:

#bba \ \ # is the coefficient of #x^2#
#bbh \ \ # is the axis of symmetry.
#bbk \ \ # is the maximum or minimum value of the function.

From the given information it is possible to form three equation of the form:

#y=a(x-h)^2+k#

and solve these simultaneously. These will involve squared quantities and messy equations, so an easier method would be to find the quadratic in the form:

#ax^2+bx+c=0#

This will be much simpler:

We know it passes through the point #(2,6)# and has a y intercept of 6. The y intercept is the constant #bbc# in the standard form we will be using. We plug in this information to form the first equation.

#6=a(2^2)+b(2)+6=>4a+2b=0 \ \ \ \ \ [1]#

We have an #x# intercept of #1#, we know that at this value of #x#, #y=0# i.e. root of equation.

Plugging in these values along with #c=6# which we reasoned above.

#0=a(1)^2+b(1)+6=>a+b=-6 \ \ \ \ \ [2]#

Solving #[1]# and #[2]# simultaneously:

Multiply #[2]# by 2 and subtract from #[1]#

#4a+2b=0#
#2a+2b=-12#

#2a=12=>a=6#

Substituting this in #[2]#

#6+b=-6=>b=-12#

So we now have all the coefficients and constants of #ax^2+bx+c=0#

#a=6#

#b=-12#

#c=6#

#:.#

#6x^2-12x+6=0#

We still need this form: #y=a(x-h)^2+k#

It can be shown that:

#h=-b/(2a) \ \ # and # \ \ \k=f(h)#

So we have:

#h=-(-12)/(2(6))=1#

#k=f(h)=f(1)=6(1)^2-12(1)+6=0#

In vertex form we get:

#y=6(x-1)^2#

This is its graph: