Answer:
1) Please find attached the graph sowing the line of symmetry
The symmetry line is a vertical line passing through (2, -9)
2) The x-intercept are (5, 0) and (-1, 0)
The y-intercept is (0, -5)
The vertex is (2, -9)
Step-by-step explanation:
The given function is;
f(x) = x² - 4·x - 5
The data values are generated as follows;
x, f(x)
-1, 0
-0.8, -1.16
-0.6, -2.24
-0.4, -3.24
-0.2, -4.16
0, -5
0.2, -5.76
0.4, -6.44
0.6, -7.04
0.8, -7.56
1, -8
1.2, -8.36
1.4, -8.64
1.6, -8.84
1.8, -8.96
2, -9
2.2, -8.96
2.4, -8.84
2.6, -8.64
2.8, -8.36
3, -8
3.2, -7.56
3.4, -7.04
3.6, 6.44
3.8, -5.76
4, -5
4.2, -4.16
4.4, -3.24
4.6, -2.24
4.8, -1.16
5, 0
The minimum is found from differentiating the function, f(x), with respect to x and looking for the zeros of the result as follows;
f'(x) = 2·x -4
f'(x) = 0 = 2·x -4
x = 2
The y-coordinate gives; f(2) = 2² - 4×2 - 5 = -9
Therefore, the symmetry line is a vertical line passing through (2, -9)
The x-intercept is the point at which y = 0, therefore, from f(x) = x² - 4·x - 5, we have;
0 = x² - 4·x - 5 = (x - 5)·(x + 1)
Therefore, the x-intercept are x = 5 or -1
The x-intercept are (5, 0) and (-1, 0)
The y-intercept occur at the point where the x value = 0, therefore, we have;
The y-intercept occur at y = f(0) = 0² - 4·0 - 5 = -5
The y-intercept is (0, -5)
Re-writing the equation in vertex form y = a(x - h)² + k gives;
f(x) = x² - 4·x - 5 = 1·(x - 2)² - 9
Therefore, the vertex is (2, -9)
