# How do you find the vertex and intercepts for #y=2(x-3)^2 +4#?

see explanation.

This has no real solutions hence f(x) has no x-intercepts. graph{2(x-3)^2+4 [-31.56, 31.67, -15.8, 15.8]}

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To find the vertex of the parabola represented by the equation ( y = 2(x - 3)^2 + 4 ), we use the formula for the vertex of a parabola in the form ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. In this equation, ( h = 3 ) and ( k = 4 ), so the vertex is at ( (3, 4) ).

To find the x-intercepts, we set ( y = 0 ) and solve for ( x ). [ 0 = 2(x - 3)^2 + 4 \implies 2(x - 3)^2 = -4 \implies (x - 3)^2 = -2 ] This equation has no real solutions because a square cannot be negative.

To find the y-intercept, we set ( x = 0 ) and solve for ( y ). [ y = 2(0 - 3)^2 + 4 \implies y = 2(9) + 4 \implies y = 18 + 4 \implies y = 22 ] So, the y-intercept is at ( (0, 22) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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