Question

for the function, the vertex of the functions graph is given. find the unknown coefficients.
y = x² + bx + c, (-2,7)

b = 4 and c = \square

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( y = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y = x^2+bx + c \), \( a = 1 \). The x - coordinate of the vertex is - 2. So we have the equation \( - 2=-\frac{b}{2\times1} \). We already know \( b = 4 \), let's verify: \( -\frac{4}{2\times1}=-2 \), which matches the x - coordinate of the vertex.

Step2: Substitute vertex into function

Now that we know the vertex is \( (-2,7) \) and the function is \( y=x^2 + bx + c \) with \( b = 4 \), we substitute \( x=-2 \), \( y = 7 \) and \( b = 4 \) into the function. So we get \( 7=(-2)^2+4\times(-2)+c \).

Step3: Solve for c

First, calculate \( (-2)^2 = 4 \) and \( 4\times(-2)=-8 \). Then the equation becomes \( 7 = 4-8 + c \). Simplify the right - hand side: \( 4-8=-4 \), so the equation is \( 7=-4 + c \). Add 4 to both sides of the equation: \( c=7 + 4=3 \).

Answer:

\( c = 3 \)