Sure! Let's solve each part of the question one by one.
### 3.1 Intersection with the y-axis
To find where the graph intersects the y-axis, we set [tex]\(x = 0\)[/tex]:
[tex]\[ g(0) = 0^2 - 5 \cdot 0 - 6 = -6 \][/tex]
Answer: The graph intersects the y-axis at [tex]\((0, -6)\)[/tex].
### 3.2 Coordinates of the x-intercepts
To find the x-intercepts, we set [tex]\(g(x) = 0\)[/tex]:
[tex]\[ x^2 - 5x - 6 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -5\)[/tex], and [tex]\(c = -6\)[/tex]:
[tex]\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{25 + 24}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm \sqrt{49}}{2} \][/tex]
[tex]\[ x = \frac{5 \pm 7}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{12}{2} = 6 \][/tex]
[tex]\[ x = \frac{-2}{2} = -1 \][/tex]
Answer: The x-intercepts are at [tex]\((6, 0)\)[/tex] and [tex]\((-1, 0)\)[/tex].
### 3.3 Equation of the axis of symmetry and turning point
The axis of symmetry for a quadratic function [tex]\(ax^2 + bx + c\)[/tex] is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For the given function [tex]\(g(x) = x^2 - 5x - 6\)[/tex]:
[tex]\[ x = -\frac{-5}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{5}{2} \][/tex]
[tex]\[ x = 2.5 \][/tex]
To find the turning point, we substitute [tex]\( x = 2.5 \)[/tex] back into the function:
[tex]\[ g(2.5) = (2.5)^2 - 5 \cdot 2.5 - 6 \][/tex]
[tex]\[ g(2.5) = 6.25 - 12.5 - 6 \][/tex]
[tex]\[ g(2.5) = -12.25 \][/tex]
Answer: The axis of symmetry is [tex]\(x = 2.5\)[/tex]. The turning point is [tex]\((2.5, -12.25)\)[/tex].
### 3.4 Range of [tex]\(g\)[/tex]
Since [tex]\(g(x) = x^2 - 5x - 6\)[/tex] is a quadratic function opening upwards (because the coefficient of [tex]\(x^2\)[/tex] is positive), the range is all values [tex]\(y \geq \)[/tex] the minimum value found at the turning point.
Answer: The range of [tex]\(g\)[/tex] is [tex]\([-12.25, \infty)\)[/tex].
### 3.5 Graph of [tex]\(f(x) = 2x - 6\)[/tex]
The equation given is [tex]\(y - 2x = -6\)[/tex]. We can rearrange it to the slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ y = 2x - 6 \][/tex]
### 3.6 Drawing the graph
Let's graph both [tex]\(g(x) = x^2 - 5x - 6\)[/tex] and [tex]\(f(x) = 2x - 6\)[/tex] on the same set of axes. (It's expected to use graphing tools or manually sketch the graphs.)
### 3.7 Intersection Points
To find where [tex]\(f(x) = g(x)\)[/tex]:
[tex]\[ x^2 - 5x - 6 = 2x - 6 \][/tex]
[tex]\[ x^2 - 5x - 6 - 2x + 6 = 0 \][/tex]
[tex]\[ x^2 - 7x = 0 \][/tex]
[tex]\[ x(x - 7) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = 0 \text{ or } x = 7 \][/tex]
Substitute these [tex]\(x\)[/tex] values into [tex]\(f(x)\)[/tex] or [tex]\(g(x)\)[/tex] to find [tex]\(y\)[/tex]:
For [tex]\(x = 0\)[/tex]:
[tex]\[ y = 2(0) - 6 = -6 \][/tex]
So, one intersection point is [tex]\((0, -6)\)[/tex].
For [tex]\(x = 7\)[/tex]:
[tex]\[ y = 2(7) - 6 = 14 - 6 = 8 \][/tex]
So, the other intersection point is [tex]\((7, 8)\)[/tex].
Answer: The points are [tex]\(A(0, -6)\)[/tex] and [tex]\(B(7, 8)\)[/tex].
### 3.8 Solving [tex]\(g(x) \leq f(x)\)[/tex]
We need to find the values of [tex]\(x\)[/tex] where:
[tex]\[ x^2 - 7x \leq 0 \][/tex]
This inequality can be solved by finding the sign changes around the roots of [tex]\(x(x - 7) = 0\)[/tex]:
The roots are [tex]\(x = 0\)[/tex] and [tex]\(x = 7\)[/tex]. Test intervals around these points:
1. [tex]\( x < 0 \)[/tex]: Choose [tex]\(x = -1\)[/tex]. [tex]\( (-1)(-8) = 8 > 0\)[/tex]
2. [tex]\( 0 < x < 7 \)[/tex]: Choose [tex]\(x = 1\)[/tex]. [tex]\( (1)(-6) = -6 < 0\)[/tex]
3. [tex]\( x > 7 \)[/tex]: Choose [tex]\(x = 8\)[/tex]. [tex]\( (8)(1) = 8 > 0\)[/tex]
Thus, [tex]\(g(x) \leq f(x)\)[/tex] in the interval:
Answer: The values of [tex]\(x\)[/tex] are [tex]\( x \in [0, 7] \)[/tex].
