Question:- On a coordinate plane, a parabola opens down. It goes through (negative 6, negative 2), has a vertex at (negative 4, 2), and goes through (negative 2, negative 2).
Answer:
Tempestt’s graph could be either a quadratic or a cubic function that has its maximum at the point (-4; 2). Let’s focus on the quadratic function as it’s simpler and more commonly used:
f(x) = – (x+4)^2 + 2
Quadratic function in this case has its vertex at (-4,2) which refers to the maximum point of the described parabola. The negative sign on the squared salary symbolises the fact that the parabolic shape faces down, or in other words, the U shape is inverted. This shape has the characteristic of having a maximum rather than a minimum as the top point of the curve.
Analyzing the components of this function, it is possible to explain why its maximum is gained at (-4, 2). The term (x+4)^ 2 transcribes the parabola 4 units to the left and the axis of symmetry is x = -4. The addition of +2 at the end of the equation shifts the whole graph of the parabola up by 2 units and the vertex is on the y-axis y = 2. Combined, these transformations relocate the vertex to (- 4, 2), which is the point with the maximum y value.
It is significant to verify this by calculating just a few points. For instance, at the critical point of x = – 4 the value of f(- 4) = – [-(4 – 4) squared + 2] = 2 as expected for the maximum y-value. In addition to this performance, no matter what value we will assign to the x-axis, if this value is less than -4 or greater than -4, the y-value will always be less than 2. For example, f(-5) = (–5+4)²–2=1 and f (-3)=(–3+4)²–2=1 (Myassignmenthelp.com, 2024).
This is a quadratic function, easy to work with and described as a graph with a maximum at (-4, 2), which is likely to be Tempestt’s graph.
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