Question

which equation could generate the curve in the graph below?

\\(y = 9x^2 + 6x + 4\\)
\\(y = 6x^2 - 12x - 6\\)
\\(y = 3x^2 + 7x + 5\\)
\\(y = 2x^2 + 8x + 8\\)

Explanation:

Analyze the graph features

Using the Graph Interpretation knowledge point, we observe:

• The curve is a parabola opening upwards, which means the leading coefficient \(a\) of the quadratic equation \(y = ax^2 + bx + c\) must be positive (\(a > 0\)).
• The vertex of the parabola lies exactly on the negative \(x\)-axis. This means the quadratic equation has exactly one real root (a double root) on the negative \(x\)-axis.
• Since there is only one real root, the quadratic must be a perfect square trinomial of the form \(y = a(x - h)^2\) where the vertex is at \((h, 0)\).
• Since the vertex is on the negative \(x\)-axis, we have \(h < 0\), which means the factored form is \(y = a(x + |h|)^2 = a(x^2 + 2|h|x + |h|^2)\). Thus, both the linear coefficient \(b\) and the constant term \(c\) must be positive.

Test the given options

Using the Perfect Square Trinomials and Factoring Quadratics knowledge points:

• Option 1: \(y = 9x^2 + 6x + 4\)

Using the Roots of Polynomials knowledge point

$$ D = b^2 - 4ac = 6^2 - 4(9)(4) = 36 - 144 = -108 < 0 $$

This has no real roots.

• Option 2: \(y = 6x^2 - 12x - 6\)

The constant term is negative, and it is not a perfect square.

• Option 3: \(y = 3x^2 + 7x + 5\)

Using the Roots of Polynomials knowledge point

$$ D = b^2 - 4ac = 7^2 - 4(3)(5) = 49 - 60 = -11 < 0 $$

This has no real roots.

• Option 4: \(y = 2x^2 + 8x + 8\)

Using the Factoring Quadratics and Perfect Square Trinomials knowledge points

$$ y = 2(x^2 + 4x + 4) = 2(x + 2)^2 $$

This is a perfect square with a single root (multiplicity 2) at \(x = -2\), which matches a vertex on the negative \(x\)-axis.

Answer:

• (A) \(y = 9x^2 + 6x + 4\)
• (B) \(y = 6x^2 - 12x - 6\)
• (C) \(y = 3x^2 + 7x + 5\)
• (D) \(y = 2x^2 + 8x + 8\) (Correct answer)