SOLUTION: Graph the function {{{ f(x)=5(x+1)^2+4 }}} Identify vertex and vertical asymptote. Thank you.

Algebra ->  Graphs -> SOLUTION: Graph the function {{{ f(x)=5(x+1)^2+4 }}} Identify vertex and vertical asymptote. Thank you.      Log On


   

Question 1103410: Graph the function +f%28x%29=5%28x%2B1%29%5E2%2B4+
Identify vertex and vertical asymptote. Thank you.
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39627) About Me  (Show Source):
You can put this solution on YOUR website!
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Identify vertex and vertical asymptote.
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No vertical asymptote for quadratic function like that one.

You can read the vertex directly from the defined function:
(-1,4) vertex.

You can also identify x-intercepts and maybe y-intercept to help in graphing. Solve for f%28x%29=0...

graph%28300%2C300%2C-6%2C2%2C-2%2C10%2C5%28x%2B1%29%5E2%2B4%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Here's the plot:
+graph%28+400%2C400%2C+-5%2C+5%2C-5%2C+15%2C+5%2A%28+x%2B1+%29%5E2+%2B+4+%29+
+f%28x%29+=+5%2A%28+x%2B1+%29%5E2+%2B+4+
+f%28x%29+=+5%2A%28+x%5E2+%2B+2x+%2B+1+%29+%2B+4+
+f%28x%29+=+5x%5E2+%2B+10x+%2B+5+%2B+4+
+f%28x%29+=+5x%5E2+%2B+10x+%2B+9++
the x-value of the vertex is:
+x%5Bv%5D+=+-b%2F%282a%29+
+a+=+5+
+b+=+10+
+x%5Bv%5D+=+%28-10%29%2F%28+2%2A5+%29+
+x%5Bv%5D+=+-1+
Plug this back into equation
+f%28x%29+=+5%2A%28+-1+%2B+1+%29+%2B+4+
+f%28x%29+=+4+
So, ther vertex is at ( -1, 4 )
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There are no asymptotes with a parabola