For each of the parabola, listed further down, do each of the following:
State the value of the \(x^2\) coefficient, \(a\) and whether it is concave-up or concave-down.
State whether its vertex is a maximum or a minimum.
State the coordinates of its \(y\)-intercept.
The parabola are listed here:
\(y=3x^2-7x+1\)
\(y = -2x^2+3x - 5\)
\(y=-x^2+4x-8\)
\(y=x^2+5x-10\)
\(y = \frac{x^2}{2}-4x+5\)
Exercise 1: Answer Key
For \(y=3x^2-7x+1\) we find:
\(a = 3\), the parabola is concave-up
Vertex is a minimum
Its \(y\)-intercept has coordinates \(\begin{pmatrix}0,1\end{pmatrix}\).
For \(y=-2x^2+3x-5\) we find:
\(a = -2\), the parabola is concave-down
Vertex is a maximum
Its \(y\)-intercept has coordinates \(\begin{pmatrix}0,-5\end{pmatrix}\).
For \(y=-x^2+4x-8\) we find:
\(a = -1\), the parabola is concave-down
Vertex is a maximum
Its \(y\)-intercept has coordinates \(\begin{pmatrix}0,-8\end{pmatrix}\).
For \(y=x^2+5x-10\) we find:
\(a = 1\) the parabola is concave-up
Vertex is a minimum
Its \(y\)-intercept has coordinates \(\begin{pmatrix}0,-10\end{pmatrix}\).
For \(y=\frac{x^2}{2}-4x+5\) we find:
\(a = \frac{1}{2}\), the parabola is concave-up
Vertex is a minimum
Its \(y\)-intercept has coordinates \(\begin{pmatrix}0,5\end{pmatrix}\).
Answers Without Working
Exercise 2
Find the equation of the axis of symmetry of each of the following parabola:
\(y=x^2 - 2x - 6\)
\(y = -x^2+6x-4\)
\(y = 2x^2-20x+18\)
\(y=4x^2-4x+1\)
\(y=\frac{x^2}{2}+x+2\)
Exercise 2: Answer Key
For \(y = x^2-2x-6\) the axis of symmetry has equation:
\[x = 1\]
For \(y = -x^2+6x-4\) the axis of symmetry has equation:
\[x = -3\]
For \(y = 2x^2-20x+18\) the axis of symmetry has equation:
\[x = 5\]
For \(y = 4x^2-4x+1\) the axis of symmetry has equation:
\[x = \frac{1}{2}\]
For \(y = \frac{x^2}{2}+x+2\) the axis of symmetry has equation:
\[x = -1\]
Exercise 3
For each of the parabola, listed further down, do each of the following:
State whether it is concave-up or concave-down.
Find the coordinates of its vertex.
State its range.
The parabola are listed here:
\(y=2x^2-12x+10\)
\(y = -x^2-6x-5\)
\(y = -3x^2+6x\)
\(y=3x^2+12x+2\)
\(y=x^2-2x+1\)
Exercise 3: Answer Key
For \(y = 2x^2-12x+10\), we find the following:
Concave-up
Vertex has coordinates \(\begin{pmatrix}3,-8\end{pmatrix}\).
The range is: \(y\geq -8\).
For \(y = -x^2-6x-5\), we find the following:
Concave-down
Vertex has coordinates \(\begin{pmatrix}-3,4\end{pmatrix}\).
The range is: \(y\leq 4\).
For \(y = -3x^2+6x\), we find the following:
Concave-down
Vertex has coordinates \(\begin{pmatrix}1,3\end{pmatrix}\).
The range is: \(y\leq 3\).
For \(y = 3x^2+12x+2\), we find the following:
Concave-up
Vertex has coordinates \(\begin{pmatrix}-2,-10\end{pmatrix}\).
The range is: \(y\geq -10\).
For \(y = x^2-2x+1\), we find the following:
Concave-up
Vertex has coordinates \(\begin{pmatrix}1,0\end{pmatrix}\).
The range is: \(y\geq 0\).
Exercise 4
For each of the following parabola find the coordinates of its \(x\)-intercepts:
\(y=x^2+x-6\)
\(y = -2x^2-4x+6\)
\(y = 7x^2-14x+7\)
\(y=4x^2+20x\)
\(y=-x^2-10x-25\)
\(y = 3x^2-9x-12\)
\(y = x^2+4x + 7\)
\(y = -9x^2+3x+2\)
\(y = -2x^2+2x-20\)
\(y = 8x^2+8x+2\)
Exercise 4: Answer Key
For \(y=x^2+x-6\) we find two \(x\)-intercepts:
\[\begin{pmatrix}-3,0\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}2,0\end{pmatrix}\]
For \(y = -2x^2-4x+6\) we find two \(x\)-intercepts:
\[\begin{pmatrix}-3,0\end{pmatrix} \quad \text{and} \quad \begin{pmatrix}1,0\end{pmatrix}\]
For \(y = 7x^2-14x+7\) we find one \(x\)-intercept:
\[\begin{pmatrix}1,0\end{pmatrix}\]
For \(y=4x^2+20x\) we find two \(x\)-intercepts:
\[\begin{pmatrix}-5,0\end{pmatrix} \quad \text{and}\quad \begin{pmatrix}0,0\end{pmatrix}\]
For \(y=-x^2-10x-25\) we find one \(x\)-intercept:
\[\begin{pmatrix}-5,0\end{pmatrix}\]
For \(y = 3x^2-9x-12\) we find two \(x\)-intercepts:
\[\begin{pmatrix}-1,0\end{pmatrix} \quad \text{and}\quad \begin{pmatrix}4,0\end{pmatrix}\]
For \(y = x^2+4x + 7\) we find:
there are no \(x\)-intercepts.
For \(y = -9x^2+3x+2\) we find two \(x\)-intercepts:
\[\begin{pmatrix}-\frac{1}{3},0\end{pmatrix} \quad \text{and}\quad \begin{pmatrix}\frac{2}{3},0\end{pmatrix}\]
For \(y = -2x^2+2x-20\) we find:
there are no \(x\)-intercepts.
For \(y = 8x^2+8x+2\) we find one \(x\)-intercept:
\[\begin{pmatrix}-\frac{1}{2},0\end{pmatrix}\]
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
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:
\[\begin{pmatrix}3,-8\end{pmatrix}\]
using all of these results, we sketch the parabola as follows:
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:
\[\begin{pmatrix}1,9\end{pmatrix}\]
using all of these results, we sketch the parabola as follows:
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:
\[\begin{pmatrix}-1,0\end{pmatrix}\]
using all of these results, we sketch the parabola as follows:
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:
\[\begin{pmatrix}4,3\end{pmatrix}\]
using all of these results, we sketch the parabola as follows:
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:
\[\begin{pmatrix}-1,-8\end{pmatrix}\]
using all of these results, we sketch the parabola as follows: