How do you re-write x squared plus 5x - 3 in vertex form?

1 Answer

Answer :

8

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Last Answer : If: y = 5x +10 and y = x^2 +4Then: x^2 +4 = 5x +10Transposing terms: x^2 -5x -6 = 0Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x =-1Therefore by substitution endpoints of the line are ... .5 = -1/5(x-2.25) => 5y= -x+114.75Perpendicular bisector equation in its general form: x+5y-114.75= 0

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Description : What are the points of intersection of the line 2x plus 5y equals 4 with the curve y squared equals x plus 4?

Last Answer : If: 2x +5y = 4 then 25y^2 = 4x^2 -16x +16If: y^2 = x +4 then 25y^2 = 25x +100So: 4x^2 -16x +16 = 25x +100Transposing terms: 4x^2 -41x -84 = 0Factorizing the above: (4x+7)(x-12) = 0 meaning x = -7/4 or x =12By substitution into original equation points of intersection:(-7/4, 3/2) and (12, -4)

Description : What are the values of x y and k when the line y equals 3x plus 1 is a tangent to the circle x squared plus y squared equals k?

Last Answer : If y = 3x + 1 is a tangent to x² + y² = k (k > 0 since it isa square), then where they meet has a repeated root; they meetat:x² + (3x + 1)² = k→ x² + 9x² + 6x + 1 - k = 0→ 10x² + 6x + (1 - k) = 0This is the ... 3/10→ y = 3 -3/10 + 1 = 9/10 + 1 = 1/10→ point of contact is (-3/10, 1/10) with k = 1/10

Description : How do ypu factor X squared minus 13x plus 12?

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Description : What is X squared plus x plus x plus one?

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Description : How do ypu factor X squared minus 13x plus 12?

Last Answer : It is x^2 -13x +12 = (x-1)(x-12) when factored

Description : What is X squared plus x plus x plus one?

Last Answer : It is the same as: x^2 +2x +1 and it is (x+1)(x+1) whenfactored

Description : What is the point of contact when the tangent line y equals x plus c touches the circle x squared plus y squared equals 4?

Last Answer : The line meets the circle when:y = x + c→ x² + y² = 4→ x² + (x + c)² - 4 = 0→ x² + x² +2cx + c² - 4 = 0→ 2x² + 2cx + (c² - 4) = 0If the line is a tangent to the circle, this has a repeatedroot. This occurs ... 2 = 0→ (x + √2)² = 0→ x = -√2→ y = x + 2√2= -√2 + 2√2= √2→ point of contact is (-√2, √2)

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Last Answer : There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos ... + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1â†' a² + b² = 89

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Description : The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

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Description : What are the points of contact when the line x -y equals 2 crosses the curve x squared -4y squared equals 5 showing work?

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