Question 146893
Find the quadratic function that has a vertex at (-1,3) and whose graph passes through the point (2,1).
<pre><font size = 4 color = "indigo"><b>
All quadratic functions are in the form

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

or for convenience, let's write {{{y}}} for {{{f(x)}}}

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

where the vertex is ({{{h}}},{{{k}}}).

So since the vertex is ({{{-1}}},{{{3}}}), we plug those in

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

{{{y = a(x-(-1))^2+3}}}

{{{y = a(x+1)^2+3}}}

Now all we need is {{{a}}}

So we substitute ({{{x}}},{{{y}}}) = ({{{2}}},{{{1}}}).

{{{y = a(x+1)^2+3}}}
{{{1 = a(2+1)^2+3}}}
{{{1 = a(3)^2+3}}}
{{{1 = 9a+3}}}
{{{-2=9a}}}
{{{-2/9=a}}}

So

{{{y = a(x+1)^2+3}}}

becomes

{{{y = -2/9}}}{{{(x+1)^2+3}}}

or

{{{f(x) = -2/9}}}{{{(x+1)^2+3}}}

in functional notation.

To draw the graph, plot the given vertex and the given
point, and find some more points, say: 

(-7,-5), (-4,1), (5,-5)

{{{drawing(400,375,-9,7,-7,7, graph(400,375,-9,7,-7,7),
locate (-7-.15,-5+.35,o),
locate (-4-.15,1+.35,o),
locate (-1-.15,3+.35,o),
locate (2-.15,1+.35,o),
locate (5-.15,-5+.35,o)
 )}}}

Now draw in the graph of the function:

{{{drawing(400,375,-9,7,-7,7, graph(400,375,-9,7,-7,7,(-2/9)(x+1)^2+3),
locate (-7-.15,-5+.35,o), 
locate (-4-.15,1+.35,o),
locate (-1-.15,3+.35,o),
locate (2-.15,1+.35,o),
locate (5-.15,-5+.35,o)
 )}}} 
Edwin</pre>