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Algebra, STD2 EQ-Bank 39

The graph of the parabola \(y=a(x+1)(x+7)\) 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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\(\text{Vertex:}\ (-4,-27)\)

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\(\text{Since graph passes through}\ (0,21):\)

\(21\) \(=a(0+1)(0+7)\)
\(21\) \(=7a\)
\(a\) \(=3\)

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

\(\Rightarrow \ x=\dfrac{-7+(-1)}{2}=-4\)

\(y\) \(=3(-4+1)(-4+7)\)
  \(=3 \times (-3) \times 3\)
  \(=-27\)

 
\(\therefore\ \text{Vertex at}\ (-4,-27).\)