To determine the correct equation for the function represented by the concave up parabola, we need to analyze the given options and consider the properties of a parabola. The general form of a parabola that opens upwards is given by:
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola, and \(a\) is a positive constant that determines how "wide" or "narrow" the parabola is.
From the options provided:
- y=(x−5)²−3 - This has vertex at (5, -3).
- y=3(x−5)²−3 - Vertex at (5, -3) but steeper due to the coefficient 3.
- y=2(x−5)²−3 - Vertex at (5, -3) and shows the parabola is also narrower than the first option.
- y=−(x−3)²−5 - This is a downward opening parabola, which doesn't fit because we need a concave up parabola.
Since we’re given that the parabola is concave up and passes through the 1st and 4th quadrants, we look for an equation that has its vertex above the x-axis (or at least the graph must be above the x-axis at points in the 1st quadrant).
Considering the possible vertices:
- If the parabola's vertex is at \( (5, -3) \), then it also opens up towards quadrants 1 and 4 if \( a > 0 \).
Among the first three options, they all have the vertex at (5, -3):
- The first one would touch (5, -3) and could be the point below the x-axis.
- The second and third options are steeper but still pass through (5, -3) and open upwards.
Without more information on specific points through which the parabola passes or any additional constraints, it’s hard to definitively select one without visualization.
However, if we're likely looking for the function that remains manageable with respect to the coordinates given, the simplest (as well as correct for basic upward openings) selection based on standard upward parabolic behavior is:
y = (x - 5)² - 3
This will yield a typical concave up parabola centered at the point (5, -3) with respect to the x-axis.
Hence, the best choice is:
y=(x−5)²−3.