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Algebra, STD2 EQ-Bank 38

The graph of the parabola \(y=a(x+2)(x-6)\) for some value of \(a\) is shown.
 

By first finding the value of \(a\), find the coordinates of the vertex.   (3 marks)

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\(\text{Vertex:}\ (2,32)\)

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\(\text{Since graph passes through}\ (0,24):\)

\(24\) \(=a(0+2)(0-6)\)
\(24\) \(=-12a\)
\(a\) \(=-2\)

 
\(\text{Vertex is halfway between \(x\)-intercepts.}\)

\(\Rightarrow \ x=\dfrac{-2+6}{2}=2\)

\(y\) \(=-2(2+2)(2-6)\)
  \(=-2 \times 4 \times (-4)\)
  \(=32\)

 
\(\therefore\ \text{Vertex at}\ (2,32).\)