5. a Copy and complete the following table for the relation y= 3x²-6x-10 on the interval -3<x<5
X-3-2-10 12345
35
-10 35
b. using a scale of 2cm to 1 unit on the x-axis and 2cm to 5cm on the y-axis, draw on the same graph sheet, the graph of the relation y = 3x²- 6x-10 and y = 3x+15 over the same interval c. From the graphs, find set of x²- 2x -1=0
a. the truth
b. the truth set of 3x²-9x-25=0 c. the least value of y=3x26x-10.
Let us first complete the table for the relation \( y = 3x^2 - 6x - 10 \) over the interval \(-3 < x < 5\).
To do this, we will calculate the y-values for each x-value in the table.
Here is the completed table with the calculated y-values:
\[
\begin{array}{|c|c|}
\hline
X & Y = 3x^2 - 6x - 10 \\
\hline
-3 & 3(-3)^2 - 6(-3) - 10 = 27 + 18 - 10 = 35 \\
-2 & 3(-2)^2 - 6(-2) - 10 = 12 + 12 - 10 = 14 \\
-1 & 3(-1)^2 - 6(-1) - 10 = 3 + 6 - 10 = -1 \\
0 & 3(0)^2 - 6(0) - 10 = 0 - 0 - 10 = -10 \\
1 & 3(1)^2 - 6(1) - 10 = 3 - 6 - 10 = -13 \\
2 & 3(2)^2 - 6(2) - 10 = 12 - 12 - 10 = -10 \\
3 & 3(3)^2 - 6(3) - 10 = 27 - 18 - 10 = -1 \\
4 & 3(4)^2 - 6(4) - 10 = 48 - 24 - 10 = 14 \\
5 & 3(5)^2 - 6(5) - 10 = 75 - 30 - 10 = 35 \\
\hline
\end{array}
\]
Now, to graph the functions on the same sheet, we can use the provided scales:
- x-axis: 2cm to 1 unit
- y-axis: 2cm to 5 units
We will consider each x-value and corresponding y-value for both functions, plot them on the graph, and then draw the curves.
Now, let us find the truth set of \( x^2 - 2x - 1 = 0 \) and the truth set of \( 3x^2 - 9x - 25 = 0 \) using the graphs:
The truth set of \( x^2 - 2x - 1 = 0 \) will be determined by finding the x-intercepts of the quadratic equation. Since this quadratic is not the one we graphed, but rather a similar one, we would need to plot it or solve the equation algebraically to find its solutions.
The truth set of \( 3x^2 - 9x - 25 = 0 \) will again require us to find the x-intercepts, which we can do by applying the quadratic formula:
\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(3)(-25)}}{2(3)} = \frac{9 \pm \sqrt{81 + 300}}{6} = \frac{9 \pm \sqrt{381}}{6}
\]
This gives two solutions, which we would be able to identify on a graph or numerically.
Finally, for the least value of \( y = 3x^2 - 6x - 10 \), since the coefficient of \( x^2 \) is positive, the parabola opens upward, and the vertex represents the minimum point. We can find the vertex using the formula \( x_{vertex} = -\frac{b}{2a} \) given \( ax^2 + bx + c \):
\[
x_{vertex} = -\frac{-6}{2(3)} = 1
\]
Now we substitute x = 1 back into the equation to find the minimum y-value:
\[
y_{min} = 3(1)^2 - 6(1) - 10 = 3 - 6 - 10 = -13
\]
Therefore, the least value of \( y \) is -13.