Answer:
My first steps is to enter the equation into desmos and see what the coordinates are which is (1, 19) 1 to the left and 19 up. The function is narrow and goes below the x axis.
Step-by-step explanation: Got it right (btw you can copy my answer and put it on edge) that way our answers aren't all the same :) bye!!
Transformation involves changing the position of a function.
The path of the ball from y = x^2 is obtained by
The function is given as:
[tex]\mathbf{y=-16x^2+32x+3}[/tex]
Factor out -16
[tex]\mathbf{y=-16(x^2-2x)+3}[/tex]
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Take the coefficient of x
[tex]\mathbf{k = -2}[/tex]
Divide by 2
[tex]\mathbf{k/2 = -1}[/tex]
Take it square
[tex]\mathbf{(k/2)^2 = 1}[/tex]
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Add and subtract the 1 in the bracket of [tex]\mathbf{y=-16(x^2-2x)+3}[/tex]
[tex]\mathbf{y=-16(x^2-2x+ 1 - 1)+3}[/tex]
Open bracket
[tex]\mathbf{y=-16(x^2-2x+ 1) + 16+3}[/tex]
[tex]\mathbf{y=-16(x^2-2x+ 1) + 19}[/tex]
Express as squares
[tex]\mathbf{y=-16(x- 1)^2 + 19}[/tex]
The above means that:
Read more about transformation at:
https://brainly.com/question/11709244
