Exercise 5
For each of the parabola, listed below, do each of the following:
- State the coordinates of the \(y\)-intercept.
- Find the coordinates of the \(x\)-intercepts.
- Find the equation of the axis of symmetry.
- Find the coordinates of the vertex.
- Sketch the parabola, labelling and "key points" and drawing the axis of symmetry.
The parabola are listed here:
- \(y=2x^2-12x+10\)
- \(y=-x^2+2x+8\)
- \(y=3x^2+6x+3\)
- \(y = x^2-8x+19\)
- \(y = 2x^2+4x - 6\)
Exercise 5: Answer Key
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For \(y=2x^2-12x+10\) we get the following results:
- \(y\)-intercept at \(\begin{pmatrix}0,10\end{pmatrix}\).
- two \(x\)-intercepts: \[\begin{pmatrix}1,0\end{pmatrix}\quad \text{and}\quad \begin{pmatrix}5,0\end{pmatrix}\]
- the axis of symmetry has equation: \[x = 3\]
- the vertex has coordinates: \[\begin{pmatrix}3,-8\end{pmatrix}\]
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using all of these results, we sketch the parabola as follows:
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For \(y=-x^2+2x+8\) we get the following results:
- \(y\)-intercept at \(\begin{pmatrix}0,8\end{pmatrix}\).
- two \(x\)-intercepts: \[\begin{pmatrix}-2,0\end{pmatrix}\quad \text{and}\quad \begin{pmatrix}4,0\end{pmatrix}\]
- the axis of symmetry has equation: \[x = 1\]
- the vertex has coordinates: \[\begin{pmatrix}1,9\end{pmatrix}\]
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using all of these results, we sketch the parabola as follows:
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For \(y=3x^2+6x+3\) we get the following results:
- \(y\)-intercept at \(\begin{pmatrix}0,3\end{pmatrix}\).
- one \(x\)-intercept: \[\begin{pmatrix}-1,0\end{pmatrix}\]
- the axis of symmetry has equation: \[x = -1\]
- the vertex has coordinates: \[\begin{pmatrix}-1,0\end{pmatrix}\]
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using all of these results, we sketch the parabola as follows:
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For \(y=x^2-8x+19\) we get the following results:
- \(y\)-intercept at \(\begin{pmatrix}0,19\end{pmatrix}\).
- no \(x\)-intercepts
- the axis of symmetry has equation: \[x = 4\]
- the vertex has coordinates: \[\begin{pmatrix}4,3\end{pmatrix}\]
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using all of these results, we sketch the parabola as follows:
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For \(y=2x^2+4x-6\) we get the following results:
- \(y\)-intercept at \(\begin{pmatrix}0,-6\end{pmatrix}\).
- two \(x\)-intercepts: \[\begin{pmatrix}-3,0\end{pmatrix}\quad \text{and}\quad \begin{pmatrix}1,0\end{pmatrix}\]
- the axis of symmetry has equation: \[x = -1\]
- the vertex has coordinates: \[\begin{pmatrix}-1,-8\end{pmatrix}\]
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using all of these results, we sketch the parabola as follows:
