Question

which statements are true about the graph of the function f(x) = 6x − 4 + x²? select two options. the vertex form of the function is f(x) = (x − 2)² + 2. the vertex of the function is (−3, −13). the axis of symmetry for the function is x = 3. the graph increases over the interval (−3, ∞ ). the function does not cross the x-axis.

Explanation:

Step1: Rewrite the function in vertex form

First, rewrite \( f(x)=x^{2}+6x - 4\) in vertex form by completing the square.
\(f(x)=x^{2}+6x - 4=(x^{2}+6x + 9)-9 - 4=(x + 3)^{2}-13\)

Step2: Analyze each option

• Option 1: The vertex form we found is \(f(x)=(x + 3)^{2}-13\), not \((x - 2)^{2}+2\), so this is false.
• Option 2: From the vertex form \(f(x)=(x + 3)^{2}-13\), the vertex is \((-3,-13)\), so this is true.
• Option 3: The axis of symmetry for a parabola in the form \(y=a(x - h)^{2}+k\) is \(x = h\). Here \(h=-3\), so the axis of symmetry is \(x=-3\), not \(x = 3\), so this is false.
• Option 4: The coefficient of \(x^{2}\) is positive (\(a = 1>0\)), so the parabola opens upwards. The vertex is at \(x=-3\), so the graph increases for \(x>-3\), i.e., over the interval \((-3,\infty)\), so this is true.
• Option 5: To check if it crosses the \(x\)-axis, we can find the discriminant of the quadratic equation \(x^{2}+6x - 4=0\). The discriminant \(D=b^{2}-4ac=6^{2}-4\times1\times(-4)=36 + 16=52>0\), so the function has two real roots and crosses the \(x\)-axis, so this is false.

Answer:

B. The vertex of the function is \((-3, -13)\)
D. The graph increases over the interval \((-3, \infty)\)