How do you find the vertex and intercepts for #y = = 5x^2 + 4x - 3#?

Answer 1

Vertex (-2/5, -19/5)

x-coordinate of vertex: #x = -b/(2a) = -4/10 = -2/5#. y-coordinate of vertex: #y(-2/5) = 5(4/25) - 4(2/5) - 3 = 4/5 - 8/5 - 3 = -19/5# To find y-intercept, make x = 0 --> y = -3 To find x-intercepts, make y = 0, and solve the quadratic equation: 5x^2 + 4x - 3 = 0 D = d^2 = b^2 - 4ac = 16 + 60 = 76 --> #d = +- sqrt76# #x = -2/5 +- (sqrt76)/5#

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Answer 2

Savannah Anderson

To find the vertex of the quadratic function y = 5x^2 + 4x - 3, use the formula: ( x = \frac{-b}{2a} ). Substitute the values of a and b from the equation to find x-coordinate of the vertex. Then, plug the x-coordinate into the original equation to find the y-coordinate.

To find the x-intercepts, set y = 0 and solve for x.

To find the y-intercept, set x = 0 and solve for y.

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Answer from HIX Tutor

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.