Given:
[tex]\begin{gathered} f(x)=2x^2-4x-6 \\ g(x)\text{ = 3x-4} \\ h(x)\text{ = }\frac{2}{x} \\ \end{gathered}[/tex]1.1.1 The equation of the reflection of g(x) in the x-axis is of the form:
The rule for the reflection in the x-axis:
[tex](x,y)\rightarrow\text{ (x,-y)}[/tex]Applying the rule:
[tex]\begin{gathered} g^{\prime}(x)\text{ = -g(x)} \\ =-(3x\text{ -4)} \\ =\text{ 4 -3x} \end{gathered}[/tex]Hence, g'(x) = 4-3x
1.1.2 The equation of the reflection of h(x) in the y-axis.
The rule for reflection in the y-axis:
[tex](x,y)\rightarrow\text{ (-x,y)}[/tex]Applying the rule:
[tex]h^{\prime}(x)\text{ = }\frac{2}{-x}[/tex]Hence, h('x) = 2/-x
1.1.3 The values of k for which :
[tex]k=2x^2-4x-6[/tex]Re-arranging:
[tex]\begin{gathered} 2x^2-4x-6-k\text{ =0} \\ 2x^2-4x\text{ -(6+k) = 0} \end{gathered}[/tex]Using the rule that for an equation to have non-real roots, the discriminant (D) must be less than zero.
[tex]\begin{gathered} D=b^2-4ac \\ (-4)^2\text{ -4(2)-(6+k) }<\text{ 0} \end{gathered}[/tex]Simplifying we have:
[tex]\begin{gathered} 16\text{ +48+8k }<\text{ 0} \\ 64\text{ + 8k }<\text{ 0} \\ 8k\text{ }<\text{ -64} \\ \text{Divide both sides by 8} \\ \frac{8k}{8}\text{ }<\frac{-64}{8} \\ k\text{ }<\text{ -8} \end{gathered}[/tex]Hence, the values of k for which the equation has non-real roots is that k must be less than -8
1.1.4 The average gradient of f(x) between x=-4 and x=0:
First, we need to find the value of f(x) at x=-4.
[tex]\begin{gathered} f(-4)=2(-4)^2\text{ -4(-4) -6} \\ =\text{ 32+16 -6} \\ =\text{ 42} \end{gathered}[/tex]Next, we must find the value of f(x) at x=0:
[tex]\begin{gathered} f(0)=2(0)^2-4(0)\text{ -6} \\ =\text{ -6} \end{gathered}[/tex]Using the average gradient formula:
[tex]\begin{gathered} \text{Average gradient = }\frac{y_B-y_A}{x_B-x_A} \\ =\text{ }\frac{-6-42}{0-(-4)} \\ =\frac{-48}{4} \\ =\text{ -12} \end{gathered}[/tex]Hence, the average gradient is -12
