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A coordinate plane's x-axis ranges from 0 to 8 and its y-axis ranges from negative 4 to 2, both by 1-unit increments. A concave up parabola is drawn passing through the 1st and 4th quadrants.

What is the equation for the function in the graph?

(1 point)
Responses

y=−(x−3)2−5
y equals negative left parenthesis x minus 3 right parenthesis squared minus 5

y=(x−5)2−3
y equals left parenthesis x minus 5 right parenthesis squared minus 3

y=2(x−5)2−3
y equals 2 left parenthesis x minus 5 right parenthesis squared minus 3

y=3(x−5)2−3
y equals 3 left parenthesis x minus 5 right parenthesis squared minus 3
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1 answer

To determine the equation of the concave up parabola, we can analyze the characteristics of the given options based on the information provided.

  1. The parabola is concave up, so we're looking for an equation in the form \( y = a(x - h)^2 + k \) with \( a > 0 \).
  2. The vertex of the parabola can give us crucial information; it might be somewhere between the x-coordinates of the points where it crosses the x-axis (or any specific points we can observe visually if included).
  3. Since the vertex generally helps us find the parabola's symmetry, we can assume by the problem statement that it likely lies somewhere around \( (5, -3) \).

Evaluating each option:

  1. \( y = -(x - 3)^2 - 5 \) — This is a downward-opening parabola (since \( a < 0 \)).
  2. \( y = (x - 5)^2 - 3 \) — This is an upward-opening parabola (since \( a > 0 \)).
  3. \( y = 2(x - 5)^2 - 3 \) — This is also an upward-opening parabola and is steeper due to \( a = 2 \).
  4. \( y = 3(x - 5)^2 - 3 \) — This is another upward-opening parabola and is even steeper as \( a = 3 \).

Given that the parabola is concave up, options 2, 3, and 4 are our candidates. Based on the vertex being at \( (5, -3) \), the simplest way to go is option 2, which provides a moderate curvature.

So, the answer is: \( y = (x - 5)^2 - 3 \).