What is the pre-image of vertex A' if the image shown on
2 answers:
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Answer:
<h2>SEE BELOW</h2>Step-by-step explanation:
<h3>to understand this</h3><h3>you need to know about:</h3>- quadratic function
- PEMDAS
vertex:(h,k)
therefore
vertex:(-1,4)
axis of symmetry:x=h
therefore
axis of symmetry:x=-1
- to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0
vertex form of quadratic equation:
- y=a(x-h)²+k
therefore
- y=a(x-(-1))²+4
- y=a(x+1)²+4
it's to notice that we don't know what a is
therefore we have to figure it out
the graph crosses y-asix at (0,3) coordinates
so,
3=a(0+1)²+4
simplify parentheses:
simplify exponent:
therefore
our vertex form of quadratic equation is
- y=-(x+1)²+4
let's simplify it to standard form
simplify square:
simplify parentheses:
simplify addition:
therefore our answer is D)y=-x²-2x+3
the domain of the function
and the range of the function is
zeroes of the function:
factor out x and -1 respectively:
group:
therefore
