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How do you write a quadratic equation with x-intercepts: -1,2 ; point: (1,4)?

Answer 1

Avery Alexander

#y=-2x^2+2x+4#

#"the x-intercepts (roots) are"#

#x=-1" and " x=2#

#rArr(x+1)" and " (x-2)" are factors"#

#"the quadratic is the product of it's factors"#

#rArry=a(x+1)(x-2)larr" a is a constant"#

#"to find a substitute " (1,4)" into the equation"#

#rArr4=a(2)(-1)#

#rArr-2a=4rArra=-2#

#rArry=-2(x+1)(x-2)larrcolor(red)" in intercept form"#

#"distributing gives"#

#y=-2(x^2-x-2)#

#rArry=-2x^2+2x+4larrcolor(red)" in standard form"# graph{-2x^2+2x+4 [-10, 10, -5, 5]}

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Answer 2

Julia Adams

To write a quadratic equation given the x-intercepts (-1, 2) and a point (1, 4), you can use the factored form of a quadratic equation:

(y = a(x - x_1)(x - x_2))

Substitute the given x-intercepts into the equation:

(y = a(x + 1)(x - 2))

Now, use the given point (1, 4) to find the value of 'a':

(4 = a(1 + 1)(1 - 2))

Solve for 'a':

(4 = a(2)(-1))

(4 = -2a)

(a = -2)

Substitute 'a' back into the equation:

(y = -2(x + 1)(x - 2))

Expanding the equation gives you the final quadratic equation:

(y = -2(x^2 - x - 2))

(y = -2x^2 + 2x + 4)

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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.