Step 1: Analyze the given information and graph. The parabola is $g(x) = ax^2 + q$. The straight line is $f(x) = mx + c$. R and S(2; 0) are the x-intercepts of $g$. T(0; 8) is the y-intercept of $g$. Graph $f$ passes through R and T. Since $g(x) = ax^2 + q$ is a parabola with its axis of symmetry at $x=0$ (the y-axis), and S(2; 0) is an x-intercept, the other x-intercept R must be symmetric to S with respect to the y-axis. Therefore, R is (-2; 0). T(0; 8) is the y-intercept of $g$. For a parabola of the form $ax^2+q$, the y-intercept is also the vertex. Since the parabola has x-intercepts and the vertex is above the x-axis, it opens downwards. 5.1: Write down the range of $g$. The vertex of $g(x) = ax^2 + q$ is at $(0, q)$. From the graph, T(0, 8) is the vertex. Since the parabola opens downwards, the maximum value of $g(x)$ is the y-coordinate of the vertex. The maximum value is $y=8$. The range of $g$ is all real numbers less than or equal to 8. $$ \text{Range of } g: \boxed{y \le 8} $$ 5.2: Write down the x-coordinate of R. As established in Step 1, due to the symmetry of $g(x) = ax^2 + q$ about the y-axis, and S(2; 0) being an x-intercept, R must be the symmetric x-intercept. $$ \text{x-coordinate of R: } \boxed{-2} $$ 5.3: Calculate the values of $a$ and $q$. From T(0; 8) being the y-intercept of $g(x) = ax^2 + q$, we substitute $x=0$ and $g(x)=8$: $$ 8 = a(0)^2 + q $$ $$ 8 = q $$ So, $q=8$. The equation for $g(x)$ is now $g(x) = ax^2 + 8$. We use an x-intercept, for example S(2; 0), to find $a$. Substitute $x=2$ and $g(x)=0$: $$ 0 = a(2)^2 + 8 $$ $$ 0 = 4a + 8 $$ $$ 4a = -8 $$ $$ a = \frac{-8}{4} $$ $$ a = -2 $$ The values are $a = -2$ and $q = 8$. $$ \boxed{a = -2, q = 8} $$ 5.4: Determine the equation of $f$. The function $f(x) = mx + c$ is a straight line that passes through R(-2; 0) and T(0; 8). First, calculate the slope $m$: $$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{0 - (-2)} = \frac{8}{2} = 4 $$ Next, find the y-intercept $c$. Since T(0; 8) is a point on the line and its x-coordinate is 0, the y-coordinate 8 is the y-intercept. So, $c = 8$. The equation of $f$ is $f(x) = 4x + 8$. $$ \boxed{f(x) = 4x + 8} $$ 5.5: Use the graphs to determine the value(s) of $x$ for which: 5.5.1: $f(x) = g(x)$ This condition means finding the x-coordinates where the graphs of $f$ and $g$ intersect. From the graph, the intersection points are R and T. R has x-coordinate -2. T has x-coordinate 0. $$ \boxed{x = -2 \text{ or } x = 0} $$ 5.5.2: $x \cdot g(x) \le 0$ We know $g(x) = -2x^2 + 8$. The x-intercepts of $g(x)$ are $x=-2$ and $x=2$. The parabola $g(x)$ is above the x-axis (i.e., $g(x) \ge 0$) for $x \in [-2, 2]$. The parabola $g(x)$ is below the x-axis (i.e., $g(x) \le 0$) for $x \in (-\infty, -2] \cup [2, \infty)$. We need $x \cdot g(x) \le 0$. This occurs when: 1. $x \le 0$ AND $g(x) \ge 0$: This means $x \in (-\infty, 0]$ and $x \in [-2, 2]$. The intersection is $x \in [-2, 0]$. 2. $x \ge 0$ AND $g(x) \le 0$: This means $x \in [0, \infty)$ and $x \in (-\infty, -2] \cup [2, \infty)$. The intersection is $x \in [2, \infty)$. Combining these two cases, the solution is $x \in [-2, 0] \cup [2, \infty)$. $$ \boxed{x \in [-2, 0] \cup [2, \infty)} $$ 5.6: The graph $h$ is obtained when $g$ is reflected along the line $y=0$. Write down the equation of $h$ in the form $h(x) = px^2 + k$. Reflecting a function $y = g(x)$ along the line $y=0$ (the x-axis) means that the new y-values are the negative of the original y-values. So, $h(x) = -g(x)$. We found $g(x) = -2x^2 + 8$. $$ h(x) = -(-2x^2 + 8) $$ $$ h(x) = 2x^2 - 8 $$ This is in the form $h(x) = px^2 + k$, where $p=2$ and $k=-8$. $$ \boxed{h(x) = 2x^2 - 8} $$ Send me the next one 📸