Question

question 5 of 10
which of the following functions best describes this graph?
(graph of a parabola opening upwards, vertex in the third quadrant, y-intercept positive)
click here for long description
a. $y = (x - 2)(x - 6)$
b. $y = x^2 + 8x + 12$
c. $y = (x - 4)(x - 4)$
d. $y = x^2 - 2x + 6$

Explanation:

Step1: Analyze the parabola's direction and vertex

The graph is a parabola opening upwards (since the coefficient of \(x^2\) is positive for all options, we check the vertex position. The vertex is in the left half (negative x - region) and above the x - axis? Wait, no, let's check the roots or the vertex form. Alternatively, let's find the roots of each option.

Step2: Find roots for each option

• Option A: \(y=(x - 2)(x - 6)\). Set \(y = 0\), then \(x=2\) or \(x = 6\). The roots are positive, but the graph's vertex is in the negative x - side, so A is wrong.
• Option B: \(y=x^{2}+8x + 12\). Factor it: \(y=(x + 2)(x+6)\). Set \(y = 0\), \(x=-2\) or \(x=-6\). The roots are negative, which matches the graph (since the parabola crosses the x - axis in the negative x - region). Let's check the vertex. The x - coordinate of the vertex of \(y=ax^{2}+bx + c\) is \(x=-\frac{b}{2a}\). For \(y=x^{2}+8x + 12\), \(a = 1\), \(b = 8\), so \(x=-\frac{8}{2\times1}=-4\). Then \(y=(-4)^{2}+8\times(-4)+12=16-32 + 12=-4\). The vertex is \((-4,-4)\), which is in the left half (negative x) and below the y - axis? Wait, the graph in the image has a vertex in the left (negative x) and the parabola opens upwards. Let's check other options.
• Option C: \(y=(x - 4)(x - 4)=(x - 4)^{2}\). The root is \(x = 4\) (double root), positive x, so C is wrong.
• Option D: \(y=x^{2}-2x + 6\). The discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times6=4 - 24=-20<0\), so no real roots (doesn't cross the x - axis), but the graph crosses the x - axis, so D is wrong.

Answer:

B. \(y = x^{2}+8x + 12\)