Question - Analysis of a Parabola in Vertex Form
Solution:
The equation provided in the image is a parabola in vertex form:\[ y = 3(x - 1)^2 + 4 \]To find the coordinates of the turning point (vertex) of the parabola, we can directly read them from the vertex form of the quadratic equation:\[ y = a(x - h)^2 + k \]where $$(h, k)$$ is the vertex of the parabola. In this equation, $$a = 3$$, $$h = 1$$, and $$k = 4$$. Thus, the coordinates of the turning point (vertex) are:\[ (h, k) = (1, 4) \]To find the x-intercepts, set $$y = 0$$ and solve for $$x$$:\[ 0 = 3(x - 1)^2 + 4 \]\[ -4 = 3(x - 1)^2 \]\[ \frac{-4}{3} = (x - 1)^2 \]Since the square of a real number is always non-negative, the equation $$(x - 1)^2 = -4/3$$ has no real solutions. Thus, there are no x-intercepts for this parabola because it doesn’t cross the x-axis.To find the y-intercept, set $$x = 0$$ and solve for $$y$$:\[ y = 3(0 - 1)^2 + 4 \]\[ y = 3(1) + 4 \]\[ y = 3 + 4 \]\[ y = 7 \]So the y-intercept is at $$y = 7$$ or at the point $$(0, 7)$$ on the graph.Lastly, the line of symmetry of a parabola is a vertical line that passes through the vertex. For the given equation, it is the line $$x = h$$, which in this case is:\[ x = 1 \]This is the line of symmetry of the graph of the given quadratic function.
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