The graphs of two functions f and g are shown on the grid below - Junior Cycle Mathematics - Question 14 - 2013

To find the roots:

1. For $f(x)$:

   $f(x) = (x + 2)^2 - 4 = 0$

   This simplifies to:

   $(x + 2)^2 = 4$

   Taking the square root gives:

   $x + 2 = extpm 2$

   Hence, the roots are:

   $x = 0, -4$

2. For $g(x)$:

   $g(x) = (x - 3)^2 - 4 = 0$

   This simplifies to:

   $(x - 3)^2 = 4$

   Taking the square root gives:

   $x - 3 = extpm 2$

   Thus, the roots are:

   $x = 1, 5$

To sketch the graph of $h$, we recognize it is a transformed version of the basic parabola:

1. The vertex form $(x - 1)^2$ indicates a vertex at $(1, -4)$.
2. Since the coefficient is positive, the parabola opens upwards.
3. We can find additional points by plugging in values around the vertex:
   • For $x = 0: h(0) = (0 - 1)^2 - 4 = 1 - 4 = -3$
   • For $x = 2: h(2) = (2 - 1)^2 - 4 = 1 - 4 = -3$
   • For $x = 3: h(3) = (3 - 1)^2 - 4 = 4 - 4 = 0$
4. Plot these points and sketch the curve.

We first rewrite the equation:

$(x - h)^2 - 2 = -x^2 - 10x + 23$

Rearranging gives:

$(x - h)^2 = -x^2 - 10x + 25$

Factoring the right-hand side:

$(x - h)^2 = (x - 5)^2$

This indicates:

$x - h = x - 5$ or $x - h = -(x - 5)$

From the first equation, we have:

$h = 5$

Thus, the value of $h$ is 5.

The standard equation for the axis of symmetry of a quadratic function in the form $ax^2 + bx + c$ is given by:

$x = -\frac{b}{2a}$

For $f(x) = x^2 - 10x + 23$, we identify $a = 1$ and $b = -10$. Thus,

$x = -\frac{-10}{2 \cdot 1} = \frac{10}{2} = 5$

Therefore, the equation of the axis of symmetry is:

$x = 5$.