Question

the graph of a quadratic function y = g(x) is shown and f(x) = 3x² + bx + c. if the graph of f(x) is reflected over the y - axis then the new graph will have the same line of symmetry as g(x). find the value of b that makes this true
a -3
b -6
c 3

Explanation:

Step1: Recall Axis of Symmetry

For a quadratic function \( f(x) = ax^2 + bx + c \), the axis of symmetry is \( x = -\frac{b}{2a} \). When reflected over the \( y \)-axis, the new function is \( f(-x)=a(-x)^2 + b(-x)+c = ax^2 - bx + c \), and its axis of symmetry is \( x=\frac{b}{2a} \). For the two axes of symmetry to be equal, \( -\frac{b}{2a}=\frac{b}{2a} \), which implies \( b = 0 \)? Wait, no, wait. Wait, the graph of \( g(x) \) has axis of symmetry? Wait, the problem says that after reflecting \( p(x) \) over \( y \)-axis, the new graph has the same line of symmetry as \( g(x) \)? Wait, first, let's find the axis of symmetry of \( g(x) \). From the graph, the vertex is at \( (-1, 8) \), so the axis of symmetry of \( g(x) \) is \( x=-1 \).

Now, \( p(x)=3x^2 + bx + c \). The axis of symmetry of \( p(x) \) is \( x = -\frac{b}{2(3)}=-\frac{b}{6} \). When we reflect \( p(x) \) over the \( y \)-axis, the new function is \( p(-x)=3(-x)^2 + b(-x)+c = 3x^2 - bx + c \). The axis of symmetry of \( p(-x) \) is \( x=\frac{b}{6} \) (since for \( y = 3x^2 - bx + c \), axis of symmetry is \( x=\frac{b}{2(3)}=\frac{b}{6} \)).

We are told that the axis of symmetry of the reflected graph (which is \( x=\frac{b}{6} \)) is the same as the axis of symmetry of \( g(x) \), which is \( x=-1 \)? Wait, no, wait. Wait, maybe I misread. Wait, the problem says: "If the graph of \( p(x) \) is reflected over the \( y \)-axis then the new graph will have the same line of symmetry as \( g(x) \)". Wait, the graph of \( g(x) \) has axis of symmetry \( x=-1 \) (since vertex is at \( (-1,8) \)). So the reflected graph of \( p(x) \) (which is \( p(-x) \)) has axis of symmetry equal to \( x=-1 \).

The axis of symmetry of \( p(-x)=3x^2 - bx + c \) is \( x=\frac{b}{6} \) (because for \( ax^2 + bx + c \), axis is \( -b/(2a) \), so here \( a=3 \), \( b=-b \) (wait, no: \( p(-x)=3x^2 - bx + c \), so the coefficient of \( x \) is \( -b \), so axis of symmetry is \( -(-b)/(2*3)=b/(6) \)). So we set \( \frac{b}{6}=-1 \)? Wait, no, wait the axis of symmetry of \( g(x) \) is \( x=-1 \), so the reflected graph's axis of symmetry is \( x=-1 \). So \( \frac{b}{6}=-1 \)? Then \( b = -6 \)? Wait, let's check again.

Wait, original \( p(x)=3x^2 + bx + c \), axis of symmetry \( x = -b/(23) = -b/6 \). Reflected over \( y \)-axis: replace \( x \) with \( -x \), so \( p(-x)=3(-x)^2 + b(-x) + c = 3x^2 - bx + c \). Axis of symmetry of \( p(-x) \) is \( x = -(-b)/(23) = b/(6) \). We need this to be equal to the axis of symmetry of \( g(x) \). From the graph, \( g(x) \) has vertex at \( (-1,8) \), so axis of symmetry \( x=-1 \). Therefore, \( b/6 = -1 \)? Wait, no, that would give \( b=-6 \). Let's check the options. Option B is -6. Let's verify.

If \( b=-6 \), then \( p(x)=3x^2 -6x + c \), axis of symmetry \( x = -(-6)/(23)=6/6=1 \). Reflected over \( y \)-axis: \( p(-x)=3x^2 +6x + c \), axis of symmetry \( x = -6/(23)=-1 \), which matches the axis of symmetry of \( g(x) \) (which is \( x=-1 \)). Yes, that works. So \( b=-6 \).

Step2: Verify the Axis of Symmetry

1. Find axis of symmetry of \( g(x) \): Vertex at \( (-1,8) \), so \( x=-1 \).
2. Axis of symmetry of \( p(x)=3x^2 + bx + c \): \( x = -b/(2*3) = -b/6 \).
3. Axis of symmetry of reflected \( p(x) \) (i.e., \( p(-x) \)): \( x = b/6 \) (since \( p(-x)=3x^2 - bx + c \), axis is \( -(-b)/(2*3)=b/6 \)).
4. Set \( b/6 = -1 \) (since \( g(x) \)'s axis is \( x=-1 \)): Solve \( b = -6 \).
5. Verify: If \( b=-6 \), \( p(x)=3x^2 -6x + c \), axis \( x=1 \). Reflected: \( p(-x)=3x^2 +6x + c \), axis \( x=-1 \), which matches \( g(…

Answer:

B. -6