Question

17 graph the parabola given by the equation below:

(f(x) = 2x^2 + 8x + 6)

Explanation:

⚡ Using what you learned: graphing quadratic functions

Step 1: Find the vertex

The \(x\)-coordinate of the vertex of a parabola in standard form \(f(x) = ax^2 + bx + c\) is given by:

$$ x = -\frac{b}{2a} $$

For \(f(x) = 2x^2 + 8x + 6\):

$$ a = 2, \quad b = 8, \quad c = 6 $$

$$ x = -\frac{8}{2(2)} = -2 $$

Substitute \(x = -2\) back into the function to find the \(y\)-coordinate:

$$ f(-2) = 2(-2)^2 + 8(-2) + 6 $$

$$ f(-2) = 2(4) - 16 + 6 $$

$$ f(-2) = 8 - 16 + 6 = -2 $$

Vertex: \((-2, -2)\)

Step 2: Find additional points

Evaluate the function at neighboring \(x\)-values to plot the curve:

For \(x = -1\):

$$ f(-1) = 2(-1)^2 + 8(-1) + 6 = 2 - 8 + 6 = 0 $$

Point: \((-1, 0)\)

For \(x = -3\) (by symmetry):

$$ f(-3) = 0 $$

Point: \((-3, 0)\)

For \(x = 0\) (\(y\)-intercept):

$$ f(0) = 6 $$

Point: \((0, 6)\)

For \(x = -4\) (by symmetry):

$$ f(-4) = 6 $$

Point: \((-4, 6)\)

Answer:

To graph the parabola, plot the following key points on the coordinate plane:

• Vertex: \((-2, -2)\)
• \(x\)-intercepts: \((-1, 0)\) and \((-3, 0)\)
• Other points: \((0, 6)\) and \((-4, 6)\)