What is the vertex of the quadratic function f(x)=3(x-5)^2+4???

Question

What is the vertex of the quadratic function f(x)=3(x-5)^2+4???

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Nho 2 weeks 2021-08-30T15:35:35+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-30T15:37:17+00:00

    Answer:

    5,4

    Step-by-step explanation:

    3(x-5)^2+4

    3(x^2-10x+25)+4

    3x^2-30x+75+4

    3x^2-30x+79

    now,

    comparing with ax^2+bx+c

    a=3

    b=-30

    c=79

    for x, -b/2a

    = -(-30)/2*3

    =30/6

    =5

    replacing x

    3x^2+bx+c

    3*5^2+(-30)*5+79

    3*25-150+79

    75+79-150

    154-150

    4

    the vertex are(x,y)

    (5,4)

    0
    2021-08-30T15:37:34+00:00

    Answer:

    Vertex: (5, 4)

    Step-by-step explanation:

    Quadratic Transformations Equation: f(x) = a(bx – h)² + k

    a – vertical shrink/stretch

    b – horizontal shrink/stretch

    h – horizontal movement left/right

    k – vertical movement up/down

    h, k – vertex

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