How do you write a quadratic equation that passes through (0,9) and vertex is (-2,5)?

1 Answer
Tony B
Mar 17, 2016

#y=(x+2)^2+5#
The other form of this is: #" "y=x^2+4x+9#

Explanation:

Standard form is:#" "y=ax^2+bx +c#

Vertex form is: #y=a(x+color(red)(d))^2+k#.................(1)

We know it passes through two points:

#P_1->(x_1,y_1)->(0,9)#

#color(brown)(P_2->(x_2,y_2)->(color(blue)(-2),color(magenta)(5)) -> Vertex)#

Known that #x_("vertex")=(-1)xxcolor(red)(d) =color(blue)(-2) # (in this case)

#=> d = +2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now we have:

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

It is also known that #y_("vertex")=color(magenta)(5)# (in this case)

Thus equation (2) becomes

#y=a(x+color(red)(2))^2+color(magenta)(5)#.................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let us assume that #a =1 # if it is not then the plot will not pass through #P_2#. As the plot does pass through this point then the completed equation is #y=(x+2)^2+5#

The equation can also be derived using the standard form of
#y=ax^2+bx+c#. Doing so would involve a lot more work.

Tony B

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