Answer:
[tex]p(x) = -5x^3 -6x^2 + 3x + 1[/tex]
Step-by-step explanation:
Given cubic function:
[tex]p(x) = ax^3 + bx^2 + cx + d[/tex]
As point (0, 1) is on the curve, substitute x = 0 into the function, set it to 1, and solve for d:
[tex]\begin{aligned} p(0) & = 1\\ \implies a(0)^3 + b(0)^2 + c(0) + d & = 1\\ \implies d & = 1 \end{aligned}[/tex]
Differentiate the function:
[tex]\begin{aligned} p(x)& = ax^3 + bx^2 + cx + d\\\implies p'(x)&=3 \cdot ax^{3-1}+2 \cdot bx^{2-1}+1 \cdot cx^{1-1}+0 \\p'(x)&=3ax^2+2bx+c\end{aligned}[/tex]
The tangent equation at the point (0, 1) is y = 3x + 1.
Therefore, the gradient of the tangent equation when x = 0 is 3.
To find the gradient of the function at a given point, substitute the x-value of that point into the differentiated function. Therefore, substitute x = 0 into the differentiated function, set it to 3, and solve for c:
[tex]\begin{aligned}p'(0) & =3 \\ \implies 3a(0)^2+2b(0)+c & =3\\ \implies c & = 3\end{aligned}[/tex]
Substitute the found values of c and d into the function:
[tex]p(x) = ax^3 + bx^2 + 3x + 1[/tex]
Substitute point (-1, -3) into the function and solve for b:
[tex]\begin{aligned}p(-1) & = -3\\\implies a(-1)^3 + b(-1)^2 + 3(-1) + 1 & = -3\\-a+b-3+1&=-3\\-a+b&=-1\\b&=a-1\end{aligned}[/tex]
To find the turning points of a function, set the differentiated function to zero and solve for x.
As there is a turning point of function p(x) when x = -1, substitute x = -1 into the differentiated function and set it to zero (remembering to substitute the found value of c = 3 into the differentiated function):
[tex]\begin{aligned} p'(-1) & =0\\\implies 3a(-1)^2+2b(-1)+3 & = 0\\3a-2b+3&=0\end{aligned}[/tex]
Substitute the found expression for b into the equation and solve for a:
[tex]\begin{aligned}3a-2b+3&=0\\\implies 3a-2(a-1)+3&=0\\3a-2a+2+3&=0\\a+5&=0\\a&=-5\end{aligned}[/tex]
Finally, substitute the found value of a into the found expression for b and solve for b:
[tex]\begin{aligned}b & = a-1\\\implies b & = -5-1\\b & = -6\end{aligned}[/tex]
Therefore:
