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y = -3(x - 1)^2 + 5 plot five points on the parabola: the vertex, two p…

Question

y = -3(x - 1)^2 + 5
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph - a - function button.

Explanation:

Step1: Find the vertex

The equation of the parabola is in vertex form \( y = a(x - h)^2 + k \), where the vertex is \((h, k)\). For \( y = -3(x - 1)^2 + 5 \), \( h = 1 \) and \( k = 5 \), so the vertex is \((1, 5)\).

Step2: Find points to the left of the vertex (x < 1)

Let's choose \( x = 0 \):
\( y = -3(0 - 1)^2 + 5 = -3(1) + 5 = -3 + 5 = 2 \), so the point is \((0, 2)\).
Let's choose \( x = -1 \):
\( y = -3(-1 - 1)^2 + 5 = -3(4) + 5 = -12 + 5 = -7 \), so the point is \((-1, -7)\).

Step3: Find points to the right of the vertex (x > 1)

Let's choose \( x = 2 \):
\( y = -3(2 - 1)^2 + 5 = -3(1) + 5 = -3 + 5 = 2 \), so the point is \((2, 2)\).
Let's choose \( x = 3 \):
\( y = -3(3 - 1)^2 + 5 = -3(4) + 5 = -12 + 5 = -7 \), so the point is \((3, -7)\).

Answer:

The five points are: Vertex \((1, 5)\), Left points \((0, 2)\), \((-1, -7)\), Right points \((2, 2)\), \((3, -7)\)