Questions: Find the focus of the parabola defined by the equation y=-1/32 x^2+1/16 x+127/32.

Solution Steps

The given equation of the parabola is

\[ y = -\frac{1}{32} x^{2} + \frac{1}{16} x + \frac{127}{32} \]

To find the vertex, we first identify the coefficients: \( a = -\frac{1}{32} \), \( b = \frac{1}{16} \), and \( c = \frac{127}{32} \). The x-coordinate of the vertex \( h \) is calculated using the formula

\[ h = -\frac{b}{2a} = -\frac{\frac{1}{16}}{2 \cdot -\frac{1}{32}} = 1.0 \]

Next, we substitute \( h \) back into the original equation to find the y-coordinate \( k \):

\[ k = y(1) = -\frac{1}{32}(1)^{2} + \frac{1}{16}(1) + \frac{127}{32} = 4.0 \]

Thus, the vertex of the parabola is at the point \( (1.0, 4.0) \).

Since the parabola opens downwards, the focus can be found using the formula

\[ \text{Focus} = (h, k + \frac{1}{4a}) \]

Calculating \( k + \frac{1}{4a} \):

\[ \frac{1}{4a} = \frac{1}{4 \cdot -\frac{1}{32}} = -4.0 \]

Thus, the y-coordinate of the focus is

\[ k + \frac{1}{4a} = 4.0 - 4.0 = -4.0 \]

Therefore, the focus of the parabola is at the point \( (1.0, -4.0) \).

Final Answer