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When graphing y = x2 - 14x + 24 and y= 20 - 5x + 31, which value would be the best minimum x- and minimum y.values for the viewing window so that all intercepts, minimums, maximums, and points of intersection are able to be seen as closely as possible?

050
0-45
-25
O-20

User OrenD
by
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1 Answer

2 votes

Final answer:

To ensure all intercepts, minimums, maximums, and points of intersection are visible on the graph of the equations y = x^2 - 14x + 24 and y = 20 - 5x + 31, the best minimum x-value for the viewing window is -5, and the best minimum y-value is -10.

Step-by-step explanation:

To ensure that all intercepts, minimums, maximums, and points of intersection are visible on the graph of the equations y = x^2 - 14x + 24 and y = 20 - 5x + 31, we need to determine the minimum x- and y-values for the viewing window. Here's how we can find these values:

  1. Find the x-intercepts of each equation by setting y = 0 and solving for x:
  • y = x^2 - 14x + 24
  • x^2 - 14x + 24 = 0
  • Factor the equation: (x - 2)(x - 12) = 0
  • x = 2 or x = 12
  • The x-intercepts are x = 2 and x = 12
Find the y-intercepts of each equation by setting x = 0:
  • y = x^2 - 14x + 24
  • y = 0^2 - 14(0) + 24 = 24
  • The y-intercept is y = 24
Find the values of y at the vertex of each equation:
  • The vertex of the equation y = x^2 - 14x + 24 occurs at x = -b / (2a), where a and b are the coefficients of x^2 and x respectively.
  • x = -(-14) / (2 * 1) = 7
  • Substitute x = 7 into the equation to find y: y = (7)^2 - 14(7) + 24 = -1
  • The vertex is (7, -1)
Determine the minimum x- and y-values for the viewing window:
  • x-min: Choose a value to the left of the x-intercepts, such as -5.
  • x-max: Choose a value to the right of the x-intercepts, such as 15.
  • y-min: Choose a value below the y-intercept, such as -10.
  • y-max: Choose a value above the vertex y-value, such as 5.
Therefore, the best minimum x-value for the viewing window is -5, and the best minimum y-value is -10.

User Michael WS
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