Vertex: Probien 1 (10 points) Find \( b \) and \( c \) so that \( y=20 x^{2}+b x+c \) has vertex \( (-1,-3) \). \( b=\square \) \( c=\square \)

To find \( b \) and \( c \) so that the vertex of the quadratic function \( y = 20x^2 + bx + c \) is at \( (-1, -3) \), we can use the vertex form of a quadratic equation or derive it from the standard form. The vertex \( (h, k) \) of a quadratic function in standard form \( ax^2 + bx + c \) can be found using the formulas: - \( h = -\frac{b}{2a} \) - \( k = f(h) \) Substituting \( a = 20 \) and the vertex \( h = -1 \): \[ -1 = -\frac{b}{2 \times 20} \] \[ -1 = -\frac{b}{40} \] Multiplying both sides by -40 gives: \[ b = 40 \] Next, to find \( c \), substitute \( x = -1 \) into the quadratic equation and set it equal to the y-coordinate of the vertex: \[ y = 20(-1)^2 + 40(-1) + c \] \[ -3 = 20(1) - 40 + c \] \[ -3 = 20 - 40 + c \] \[ -3 = -20 + c \] Adding 20 to both sides: \[ c = 17 \] Thus, the final values are: \( b = 40 \) and \( c = 17 \).