Let's address each part of the question step-by-step.
### (a) Find the x-coordinate of the vertex of the graph:
To find the x-coordinate of the vertex for the quadratic function [tex]\(y = 3x^2 - 60x + 140\)[/tex], we use the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, the coefficients are:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = -60 \][/tex]
So, the x-coordinate of the vertex is:
[tex]\[ x = -\frac{-60}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{60}{6} \][/tex]
[tex]\[ x = 10 \][/tex]
Therefore, the x-coordinate of the vertex is [tex]\( x = 10 \)[/tex].
### (b) Set the viewing window so that the x-coordinate of the vertex is near the center of the window and the vertex is visible:
To ensure the x-coordinate of the vertex (which we found to be [tex]\(x = 10\)[/tex]) is near the center of our viewing window, we can choose an x-range that places 10 in the middle. For instance, a reasonable viewing window might be:
- [tex]\( x \)[/tex] ranging from 5 to 15,
- [tex]\( y \)[/tex] ranging from -200 to 200 (since the vertex has a y-coordinate of -160 which we computed earlier).
This way, [tex]\(x = 10\)[/tex] will be centered and the important features, including the vertex, will be visible.
### (c) State the coordinates of the vertex.
To find the y-coordinate of the vertex, we plug [tex]\(x = 10\)[/tex] back into the equation [tex]\(y = 3x^2 - 60x + 140\)[/tex]:
[tex]\[ y = 3(10)^2 - 60 \cdot 10 + 140 \][/tex]
[tex]\[ y = 3 \cdot 100 - 600 + 140 \][/tex]
[tex]\[ y = 300 - 600 + 140 \][/tex]
[tex]\[ y = -160 \][/tex]
Therefore, the coordinates of the vertex are [tex]\( (10, -160) \)[/tex].
To summarize:
- (a) The x-coordinate of the vertex is [tex]\( x = 10 \)[/tex].
- (b) Set the viewing window with an x-range from 5 to 15 and a y-range from -200 to 200.
- (c) The coordinates of the vertex are [tex]\( (10, -160) \)[/tex].
