Quadratic Functions & Parabola

Solution

At the y-intercept \(x=0\) so replacing \(x\) by \(0\) in the parabola's equation: \[\begin{aligned} y & = 0^2 + 2\times 0 - 3\\ & = 0 + 0 - 3 \\ y &= -3 \end{aligned}\] So the y-intercept is the point with coordinates \(\begin{pmatrix}0,-3\end{pmatrix}\).

Note: you don't need to show the working when you replace \(x\) by \(0\) and can state the answer directly.

The y-intercept can be seen on this parabola's graph:

Solution

Comparing \(y=x^2-6x+5\) to the generic \(y=ax^2+bx+c\), we see that \(a = 1\), \(b=-6\) and \(c=5\). Using these values and the formula for the axis of symmetry we find: \[\begin{aligned} x & = \frac{-b}{2a} \\ & = \frac{-(-6)}{2\times 1} \\ & = \frac{6}{2} \\ x & = 3 \end{aligned}\] This parabola's vertical axis of symmetry has equation: \[x = 3\] This is illustrated in the graph we see here, where the axis of symmetry is the dotted line.

Solution

1. The \(x^2\) coefficient is \(2\). Since \(2>0\) this parabola's vertex is a minimum point.
2. To find the coordinates of the vertex, we follow the two steps we read further-up:
   • Step 1: we calculate the \(x\)-coordinate, \(h\), of the vertex using the formula: \[h = \frac{-b}{2a}\] Looking at \(y=2x^2 - 4x-6\), we see that: \[a = 2, \ b = -4, \ c = -6\] So the formula for the \(x\)-coordinate becomes: \[\begin{aligned} h & = \frac{-b}{2a}\\ & = \frac{-(-4)}{2\times 2} \\ & = \frac{4}{4} \\ h & = 1 \end{aligned}\] So the \(x\)-coordinate of the vertex is \(h=1\).
   • Step 2: we calculate the \(y\)-coordinate of the vertex by replacing \(x\) by \(1\) inside \(y=2x^2-4x-6\) and calculating the value of \(y\).
     
     That's: \[\begin{aligned} y & = 2\times 1^2-4\times 1-6 \\ & = 2\times 1 - 4 - 6 \\ & = 2 - 4 - 6 \\ & = -2 - 6 \\ y & = -8 \end{aligned}\] So the \(y\)-coordinate of the vertex is \(k=-8\).
   Finally we can state that this parabola has a minimum point with coordinates \(\begin{pmatrix}1, -8 \end{pmatrix}\).

Solution

The parabola \(y = 3x^2 - 3x-6\) cuts the \(x\)-axis when \[3x^2-3x-6 = 0\] We solve this equation in two steps

• Step 1: calculate the discriminant \(\Delta \): \[\begin{aligned} \Delta & = b^2 - 4ac\\ & = (-3)^2-4\times 3\times (-6)\\ & = 9 - 12 \times (-6) \\ & = 9 - (-72)\\ & = 9 + 72 \\ \Delta & = 81 \end{aligned}\]
• Step 2: we now solve the equation, according to the sign of \(\Delta\).
  
  Since \(\Delta >0\) the equation has two solutions, given by the formula: \[x = \frac{-b\pm \sqrt{\Delta}}{2a}\] replacing \(\Delta\), \(a\) and \(b\) by their respective values leads to: \[\begin{aligned}x &= \frac{-b\pm \sqrt{\Delta}}{2a} \\ & = \frac{-(-3) \pm \sqrt{81}}{2\times 3} \\ & = \frac{3 \pm 9}{6} \\ x & = \frac{1 \pm 3}{2} \\ \end{aligned}\] So the equation \(3x^2 - 3x-6 = 0\) has two solutions:
  \[\begin{aligned} x & = \frac{1 - 3}{2} \\ & = \frac{-2}{2} \\ x & = -1 \end{aligned}\]

  \[\begin{aligned} x & = \frac{1 + 3}{2} \\ & = \frac{4}{2} \\ x & = 2 \end{aligned}\]

This means that the parabola cuts the \(x\)-axis twice, at: \[x = \begin{pmatrix}-1,0\end{pmatrix} \quad \text{and} \quad x = \begin{pmatrix}2,0\end{pmatrix} \] These two points are known as the parabola's \(x\)-intercepts.
Looking at the parabola \(y = 3x^2 - 3x-6\) on an \(xy\)-grid we can confirm these results: