For x values 1 unit either side of the vertex, the value of y is 2 more than it is at the vertex.
a = 2

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More formally, The graph appears to go through the point (4, 1). Substituting these values for (x, y) in the equation, we get
1 = a(4-3)² -1 . . . . substitute for x and y
2 = a . . . . . . . . . . add 1 and simplify

7 0

3 years ago

I need help with before 9

Westkost [7]

Answer:

There is no 9 on the paper.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A sapling that started out 14 inches tall grew 4 inches per year. How old was the sapling when it was 46 inches tall?

tatyana61 [14]

Answer:

8 years old

Step-by-step explanation:

46 - 14 = 32 inches of growth

32 inches / 4 inches = 8 years

3 0

3 years ago

N=4; 2i and 3i are zeros <br> f(-1)=50

Nonamiya [84]

Solution:- $\text{Let f(x) be any nth degree polynomial with n=4}\\$

$\text{Given that 2i and 3i are the zeroes of f(x)}$

$\text{so (x-2i) and (x-3i) are factors of f(x)}$

$\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}$

$\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)$

6 0

3 years ago