The equation of parabola is y = 3x²- 12x+ 15.
<u>Step-by-step explanation:</u>
The equation of the parabola is given as y = x².
If it has a vertex then the equation can be written as,
y = a(x-h)² + k
where (h, k ) is the vertex and a is the stretching or compressing factor.
So here the vertex is (2,3) and it is vertically stretched by 3.
So the equation is given as,
y = 3(x-2)² + 3
y = 3(x² -4x+4) + 3
= 3x² - 12x + 12 + 3
y = 3x²- 12x+ 15 is the equation of the parabola.
Answer:
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Step-by-step explanation:
Given that M is a polynomial of degree 3.
So, it has three zeros.
Let the polynomial be
M(x) =a(x-p)(x-q)(x-r)
The two zeros of the polynomial are -4 and 4i.
Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.
Then,
M(x)= a{x-(-4)}(x-4i){x-(-4i)}
=a(x+4)(x-4i)(x+4i)
=a(x+4){x²-(4i)²} [ applying the formula (a+b)(a-b)=a²-b²]
=a(x+4)(x²-16i²)
=a(x+4)(x²+16) [∵i² = -1]
=a(x³+4x²+16x+64)
Again given that M(0)= 53.12 . Putting x=0 in the polynomial
53.12 =a(0+4.0+16.0+64)

=0.83
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
It’s either 11.25 or 27.75. Next time please show the answer choices too!