How do you find the vertex and intercepts for $y = 2 {(x - 3)}^{2} + 1$ $y = 2(x - 3)^2 + 1$ ?

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Answer 1 · 2018-04-01T04:54:28

vertex is $(3, 1)$ $(3,1)$

Y intercept 19
and
No x intercept

In vertex form $f (x) = A {(B [x - C])}^{2} + D$ $f(x)=A(B[x-C])^2+D$

We know that C is the x co-ordinate of the vertex and D is the y co-ordinate

So the vertex is $(3, 1)$ $(3,1)$

Y intercept (when x 0)

$y = 2 {((0) - 3)}^{2} + 1 = 2 {(- 3)}^{2} + 1 = 18 + 1 = 19$ $y=2((0)-3)^2+1=2(-3)^2+1=18+1=19$

X intercept (when y 0)

$0 = 2 {(x - 3)}^{2} + 1$ $0=2(x-3)^2+1$
$- 1 = 2 {(x - 3)}^{2}$ $-1=2(x-3)^2$
$\sqrt{- 1} = 2 (x - 3)$ $sqrt(-1)=2(x-3)$

Root 1 doesn't exist on the number line showing that there is no x intercept

How do you find the vertex and intercepts for y=2(x−3)2+1y = 2(x - 3)^2 + 1?