Answer:
y = x² - 8x + 17
Step-by-step explanation:
Recall this formula:
y = a(x-h)² + k
Original equation was y = x², which in this form looks like
y = 1(x - 0)² + 0
If the shape of the graph didn't change, that means that a (the compressor/intensifier) stayed the same (a - value) meaning we can write the equation as..
y = (x-h)² + k
Remember, (h, k) is the vertex, which looking from your graph is at (4, 1)
y = (x - 4)² + 1
y = x^2- 8x + 16+1
y = x² - 8x + 17
-3 it’s the rise over run I believe & because it’s going down it’s negative
Answer:
Step-by-step explanation:
Writing the description in algebraic translation

so we have to find the expression which will be equal to
.
Considering the expression



so the expression becomes
∵ 


Therefore,
