+0

# Algebra

0
3
1
+1624

The graph of \$y = ax^2 + bx + c\$ has an axis of symmetry of \$x = -1.\$ If \$a = 2,\$ then find \$b.\$

May 12, 2024

#1
+269
0

The axis of symmetry, x = -1, tells us that the vertex of the parabola is located at the point (-1, y-value). In any parabola, the axis of symmetry passes exactly through the vertex.

Since we are given that a = 2, we can rewrite the equation as:

y = 2x^2 + bx + c

We know that the x-coordinate of the vertex is -1. When we plug this value for x in the equation, we should get the y-coordinate of the vertex.

However, we are not given the specific y-value of the vertex. This doesn't prevent us from finding b though!

Let's substitute x = -1 into the equation:

y = 2(-1)^2 + b(-1) + c

This simplifies to:

y = 2 - b + c

Key Point: Regardless of the specific value of c (which determines the vertical positioning of the parabola), the axis of symmetry will still pass through the vertex. This means that even though we don't know the exact y-value of the vertex, we know it must be the same on both sides of the axis of symmetry (x = -1).

Therefore, the equation when we plug in x = 1 (on the other side of the axis of symmetry) must also equal the expression we obtained above.

In other words:

y = 2(1)^2 + b(1) + c must also equal y = 2 - b + c (from above)

Simplifying the left side of the new equation:

y = 2 + b + c

Equating both sides:

2 + b + c = 2 - b + c since they represent the same y-value

Solving for b, we get:

2b = 0

Therefore, b = 0.

So, even without knowing the specific y-value of the vertex, we were able to find that b = 0.

May 13, 2024