# How do you graph the parabola #y = 4x^2 + 8x -22# using vertex, intercepts and additional points?

UNDERSTAND THE EXPLANATION

Taking out 4 as a common factor as 'a' is always a factor to be taken out

use completing the square method

I hope now can get vertex, x, and y-intercept AND axis of symmetry if not then mention me in the comment for further help

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To graph the parabola (y = 4x^2 + 8x - 22):

**Vertex**: Use the formula (x = \frac{-b}{2a}) to find the x-coordinate of the vertex. Then substitute this value into the equation to find the y-coordinate.**Intercepts**: Find the x-intercepts by setting (y = 0) and solving for (x). Find the y-intercept by setting (x = 0) and solving for (y).**Additional Points**: Choose additional values of (x) and substitute them into the equation to find corresponding (y) values.

Once you have these points, plot them on a coordinate plane and draw a smooth curve through them to represent 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.

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