Question

which statements are true about the graph of the function ( f(x) = 6x - 4 + x^2 )? 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, infty) ).
• the function does not cross the ( x )-axis.

Explanation:

Step1: Rewrite function in standard form

$f(x) = x^2 + 6x - 4$

Step2: Convert to vertex form (complete square)

$f(x) = (x^2 + 6x + 9) - 9 - 4 = (x+3)^2 -13$

Step3: Identify vertex and axis of symmetry

Vertex: $(-3, -13)$, Axis: $x=-3$

Step4: Analyze increasing interval

Since $a=1>0$, parabola opens up. Increases for $x > -3$, i.e., $(-3, \infty)$

Step5: Check x-intercepts (discriminant)

$\Delta = 6^2 - 4(1)(-4) = 36 +16=52>0$, so crosses x-axis.

Step6: Verify all options

Eliminate incorrect options, select valid ones.

Answer:

The axis of symmetry for the function is $x = -3$.
The graph increases over the interval $(-3, \infty)$.