Question

algebra 2a semester online practice
complete this assessment to review what youve learned. it will not count toward y...
43 of 48
what is the point of inflection of the new function obtained by shifting the function,
g(x)=x³ - 15, 4 units upwards from the inflection point of g(x)? (1 point)
(11,0)
(0, - 19)
(-11,0)
(0, - 11)

Explanation:

Step1: Find the inflection point of \(g(x)=x^{3}-15\)

The second - derivative of \(y = f(x)\) is used to find inflection points. For \(y=x^{3}-15\), the first derivative \(y^{\prime}=3x^{2}\), and the second derivative \(y^{\prime\prime} = 6x\). Set \(y^{\prime\prime}=0\), then \(6x = 0\), which gives \(x = 0\). When \(x = 0\), \(y=0^{3}-15=-15\). So the inflection point of \(g(x)\) is \((0,-15)\).

Step2: Shift the inflection point 4 units upwards

When a point \((x,y)\) is shifted \(k\) units upwards, the new point is \((x,y + k)\). Here \(x = 0\), \(y=-15\) and \(k = 4\). So the new \(y\) - coordinate is \(y=-15 + 4=-11\), and the \(x\) - coordinate remains \(x = 0\).

Answer:

\((0,-11)\)