Question 184530
Remember, the general vertex form of an equation is {{{y=a(x-h)^2+k}}} where "a" is the stretch/compression factor and (h,k) is the vertex. We know what the vertex is, but we don't know the value of "a". So the goal is to find "a".




Since the vertex is (-2,4), this tells us that {{{h=-2}}} and {{{k=4}}}. 



{{{y=a(x-h)^2+k}}} Start with the vertex form of the equation



{{{y=a(x-(-2))^2+4}}} Plug in {{{h=-2}}} and {{{k=4}}}



{{{y=a(x+2)^2+4}}} Rewrite {{{x-(-2)}}} as {{{x+2}}}




Now because we want the parabola to pass through the point (-1,8), this means that when {{{x=-1}}}, {{{y=8}}}. So we can plug in {{{x=-1}}} and {{{y=8}}}



{{{8=a(-1+2)^2+4}}} Plug in {{{x=-1}}} and {{{y=8}}}. We finally have an equation with one unknown variable



{{{8-4=a(-1+2)^2}}} Subtract 4 from both sides.



{{{4=a(1)^2}}} Combine like terms.



{{{4=a(1)}}} Square 1 to get 1



{{{4=a}}} Multiply



So the value of "a" is {{{a=4}}}



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Answer:


This means that the equation with a vertex of (-2,4) that passes through (-1,8) is {{{y=4(x+2)^2+4}}}



Here's the graph to visually verify the answer


{{{drawing(500,500,-5,5,-5,10,
grid(1),
graph(500,500,-5,5,-5,10,4(x+2)^2+4),
circle(-2,4,0.05),circle(-2,4,0.08),circle(-2,4,0.10),
circle(-1,8,0.05),circle(-1,8,0.08),circle(-1,8,0.10)
)}}} Graph of {{{y=4(x+2)^2+4}}} with a vertex of (-2,4) that passes through the point (-1,8)