Question

y = 2x² + 20x + 48 (vertex point a)
vertex of y = -x² + 8x + 3 (point b)
determine the equation of the perpendicular bisector of ab

Explanation:

Step1: Find vertex of \( y = 2x^2 + 20x + 48 \)

For a quadratic \( y = ax^2 + bx + c \), vertex x - coordinate is \( x = -\frac{b}{2a} \). Here, \( a = 2 \), \( b = 20 \), so \( x = -\frac{20}{2\times2} = -5 \).
Substitute \( x = -5 \) into the equation: \( y = 2(-5)^2 + 20(-5) + 48 = 2\times25 - 100 + 48 = 50 - 100 + 48 = -2 \). So vertex A is \( (-5, -2) \).

Step2: Find vertex of \( y = -x^2 + 8x + 3 \)

Using \( x = -\frac{b}{2a} \), \( a = -1 \), \( b = 8 \), so \( x = -\frac{8}{2\times(-1)} = 4 \).
Substitute \( x = 4 \): \( y = -(4)^2 + 8(4) + 3 = -16 + 32 + 3 = 19 \). So vertex B is \( (4, 19) \).

Step3: Find slope of AB

Slope \( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{19 - (-2)}{4 - (-5)} = \frac{21}{9} = \frac{7}{3} \).

Step4: Find slope of perpendicular line

Perpendicular slope \( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{3}{7} \).

Step5: Use point - slope form (using point A, for example)

Point - slope: \( y - y_1 = m(x - x_1) \). Using A \( (-5, -2) \):
\( y - (-2) = -\frac{3}{7}(x - (-5)) \)
\( y + 2 = -\frac{3}{7}(x + 5) \)
Multiply through by 7: \( 7y + 14 = -3x - 15 \)
\( 3x + 7y + 29 = 0 \) (or slope - intercept: \( y = -\frac{3}{7}x - \frac{29}{7} \))

Answer:

The equation of the perpendicular bisector of \( \overline{AB} \) is \( 3x + 7y + 29 = 0 \) (or \( y = -\frac{3}{7}x - \frac{29}{7} \))