Question

Encontre as coordenadas do vértice para cada função quadrática:
a) y = – x2 + 3x e) y = x2 + 5x + 6
b) y =– x2 + 49 f) y = x2 – 7x + 12
c) y = 2x2 – 3x – 2 g) y = 3x2 – 6x
d) y = – x2 – 2x + 3 h) y = – x2 + 6x – 5

Answer 1

                    Funções Quadráticas

Se tivermos uma função quadrática f (x) = ax² + bx + c, o vértice (h; k) é dado por:

• $\boxed{\bf{h=\dfrac{-b}{2a} }}$

• $\boxed{\bold{k=f(h)}}$

Leve em consideração que "y" é o mesmo que dizer f (x):

a) y = – x² + 3x

$h=\dfrac{-3}{2(-1)} \\\\\\h=\dfrac{-3}{-2} =\boxed{\bold{\dfrac{3}{2} }}\\\\\\\\f(\dfrac{3}{2} )=-(\dfrac{3}{2} )^{2} +3(\dfrac{3}{2} )\\\\\\f(\dfrac{3}{2} )=-\dfrac{9}{4} +\dfrac{9}{2} \\\\\\f(\dfrac{3}{2})=-\dfrac{9}{4} +\dfrac{9*2}{2*2} \\\\\\f(\dfrac{3}{2})=-\dfrac{9}{4} +\dfrac{18}{4} \\\\\\f(\dfrac{3}{2})=-\dfrac{9+18}{4} \\\\\\f(\dfrac{3}{2})=\boxed{\bold{-\dfrac{9}{4}=k}}$

Vertice (3/2 ; -9/4)

b) y = -x² + 49

$h=\dfrac{-0}{2(1)} \\\\\\h=\boxed{\bold{0 }}\\\\\\\\f(0 )=-(0 )^{2} +49)\\\\\\f(0 )=-0 +49 \\\\\\\\f(0)=\boxed{\bold{49=k}}$

Vertice (0 ; 49)

c) y = 2x² - 3x - 2

$h=\dfrac{-(-3)}{2(2)} \\\\\\h=\boxed{\bold{\dfrac{3}{4} }}\\\\\\\\f(\dfrac{3}{4} )=2(\dfrac{3}{4} )^{2} -3(\dfrac{3}{4} )-2\\\\\\f(\dfrac{3}{4} )=2(\dfrac{9}{16}) -3(\dfrac{3}{4} )-2\\\\\\f(\dfrac{3}{4})=\dfrac{18}{16} -\dfrac{9}{4} -\dfrac{2}{1} \\\\\\f(\dfrac{3}{4} )=\dfrac{18}{16} -\dfrac{9*4}{4*4} -\dfrac{2*16}{1*16} \\\\\\f(\dfrac{3}{4} )=\dfrac{18-36-32}{16} \\\\\\f(\dfrac{3}{4} )=\dfrac{-50}{16}\\\\\\f(\dfrac{3}{4} )=\boxed{\bf{\dfrac{-25}{4}=k}}$

Vertice (3/4 ; -25/4)

d) y = -x² - 2x + 3

$h=\dfrac{-(-2)}{2(-1)} \\\\\\h=\dfrac{2}{-2} \\\\\\h=\boxed{\bold{-1} }}\\\\\\\\f(-1 )=-(-1 )^{2} -2(-1 )+3\\\\\\f(-1)=-1+2+3\\\\\\f(-1)=\boxed{\bold{4=k}}$

Vertice (-1 ; 4)

e) y = x² + 5x + 6

$h=\dfrac{-5}{2(1)} \\\\\\h=\boxed{\bold{\dfrac{-5}{2} } }}\\\\\\\\f(\dfrac{-5}{2} )=(\dfrac{-5}{2} )^{2} +5(\dfrac{-5}{2} )+6\\\\\\f(-1)=\dfrac{25}{4} -\dfrac{25}{2} +6\\\\\\f(-2,5 )=6,25-12,5+6\\\\\\f(-2,5)=\boxed{\bold{-0,25=k}}$

Vertice (-2,5 ; 0,25)

f) y = x² – 7x + 12

$h=\dfrac{-(-7)}{2(1)} \\\\\\h=\dfrac{7}{2} \\\\\\\boxed{\bold{h=3,5}}\\\\\\\\f(3,5)=(3,5)^{2} -7(3,5)+12\\\\\\f(3,5)=12,25-24,5+12\\\\\\f(3,5)=\boxed{\bold{-0,25=k}}$

Vertice (3,5 ; -0,25)

g) y = 3x² – 6x

$h=\dfrac{-(-6)}{2(3)} \\\\\\h=\dfrac{6}{6} \\\\\\\boxed{\bf{h=1}}\\\\\\\\f(1)=3(1)^{2} -6(1)\\\\\\f(1)=3(1)-6\\\\\\f(1)=\boxed{\bf{-3=k}}$

Vertice (1 ; -3)

h) y = -x² + 6x - 5

$h=\dfrac{-6)}{2(-1)} \\\\\\h=\dfrac{-6}{-2} \\\\\\\boxed{\bf{h=3}}\\\\\\\\f(3)=-3^{2} +6(3)-5\\\\\\f(3)=-9+18-5\\\\\\f(3)=\boxed{\bf{4=k}}$

Vertice (3 ; 4)

Espero ter ajudado, boa sorte!!