How do you find the intercept and vertex of #y= 3(x1)²1#?
Vertex
No
This is the Vertex Form of a quadratic equation so you can virtually directly read off the coordinates of the vertex.
As the determinant is negative the curve does not cross the xaxis nor is the axis a tangent to the curve.
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To find the intercept and vertex of ( y = 3(x1)^2  1 ):

Intercept: Substitute ( x = 0 ) into the equation and solve for ( y ).
( y = 3(01)^2  1 )

Vertex: Use the formula ( x = \frac{b}{2a} ) to find the xcoordinate of the vertex. Then substitute this xvalue into the equation to find the ycoordinate.
( x = \frac{(1)}{2(3)} )
( y = 3\left(\frac{1}{2}\right)^2  1 )
( y = \frac{5}{2} )
So, the intercept is at (0, 2) and the vertex is at (1, 2.5).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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