How do you find the vertex, the y-intercept, and symmetric point, and use these to sketch the graph given #y=5x^2-30x+31#?
To do this we will need to perform some algebraic manipulation.
Symmetric Point A quadratic equation such as this will produce a parabola, which is always symmetric about its vertex. This means that the symmetric point will be the vertex.
After you have done this, it is reasonable to sketch the parabola.
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To find the vertex of the parabola represented by the equation y = 5x^2 - 30x + 31, you can use the formula for the x-coordinate of the vertex, which is given by x = -b/(2a) where a and b are the coefficients of x^2 and x respectively. Then substitute this x-value into the equation to find the corresponding y-value.
The y-intercept is found by setting x = 0 in the equation and solving for y.
The symmetric point is the point on the parabola that lies on the same horizontal line as the vertex but on the opposite side of the axis of symmetry.
Once you have found the vertex, y-intercept, and symmetric point, you can sketch the graph by plotting these points and noting the shape of the parabola.
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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.
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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