Analysis of a Parabola in Vertex Form
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.
