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

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17-09-2020
Mathematics

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

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Space Space · Answer 1 · 2020-09-17T21:29:22+02:00

Space Space

17-09-2020

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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bantawahenhse bantawahenhse · Answer 2 · 2020-09-17T21:35:45+02:00

bantawahenhse bantawahenhse

17-09-2020

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)

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