How do you find the quadratic function with vertex (3,4) and point (1,2)?

A quadratic function can be written in vertex form as:

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

where #(h,k)# is the vertex and #a# is a constant multiplier.

In our example the vertex #(h, k)# is #(color(red)(3),color(blue)(4))#, so we can write:

#f(x) = a(x-color(red)(3))^2+color(blue)(4)#

Given that this passes through the point #(color(red)(1), color(blue)(2))#, we must have:

#color(blue)(2) = a(color(red)(1)-3)^2+4 = 4a+4#

Subtract #4# from both ends to get:

#-2 = 4a#

Divide both sides by #4# and transpose to find:

#a = -1/2#

So our quadratic function can be written in vertex form as:

#f(x) = -1/2(x-3)^2+4#

We can multiply this out and simplify as follows:

#f(x) = -1/2(x-3)^2+4#

#color(white)(f(x)) = -1/2(x^2-6x+9)+4#

#color(white)(f(x)) = -1/2x^2+3x-9/2+4#

#color(white)(f(x)) = -1/2x^2+3x-1/2#

The equation of a parabola in #color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where (h ,k) are the coordinates of the vertex and a is a constant.

here the vertex = (3 ,4) #rArrh=3,k=4#

Thus we can write #y=a(x-3)^2+4#

To find a, substitute the point (1 ,2) into the equation and solve for a

#x=1 : y=2rArr2=a(1-3)^2+4#

#rArr2=4a+4rArr4a=-2rArra=-1/2#

#"Thus " y=-1/2(x-3)^2+4larrcolor(blue)" in vertex form"#

distribute and simplify gives the equation in a different form.

#y=-1/2(x^2-6x+9)+4#

#rArry=-1/2x^2+3x-9/2+4#

#y=-1/2x^2+3x-1/2larrcolor(blue)" in standard form"#