Search
- Filter Results
- Location
- Classification
- Include attachments
- https://math.libretexts.org/Courses/Penn_State_University_Greater_Allegheny/MATH_110%3A_Techniques_of_Calculus_I_(Gaydos)/01%3A_Review/1.05%3A_QuadraticsFor the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[...For the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[0=(3x-1)(x+2),\nonumber \] so either \[ \begin{align*} 0 & = 3x-1\\ x & = \frac{1}{3} \end{align*} \nonumber \] or \[ \begin{align*} 0 & = x+2\\ x & = -2 \end{align*} \nonumber \] So the Horizontal intercepts are at \( \left(\frac{1}{3},0\right) \) and \((-2,0)\).
- https://math.libretexts.org/Courses/Prince_Georges_Community_College/MAT_2160%3A_Applied_Calculus_I/06%3A_Algebra_Review/6.07%3A_QuadraticsFor the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[...For the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[0=(3x-1)(x+2),\nonumber \] so either \[ \begin{align*} 0 & = 3x-1\\ x & = \frac{1}{3} \end{align*} \nonumber \] or \[ \begin{align*} 0 & = x+2\\ x & = -2 \end{align*} \nonumber \] So the Horizontal intercepts are at \( \left(\frac{1}{3},0\right) \) and \((-2,0)\).
- https://math.libretexts.org/Courses/Butler_Community_College/MA148%3A_Calculus_with_Applications_-_Butler_CC/01%3A_Review/1.05%3A_QuadraticsFor the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[...For the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[0=(3x-1)(x+2),\nonumber \] so either \[ \begin{align*} 0 & = 3x-1\\ x & = \frac{1}{3} \end{align*} \nonumber \] or \[ \begin{align*} 0 & = x+2\\ x & = -2 \end{align*} \nonumber \] So the Horizontal intercepts are at \( \left(\frac{1}{3},0\right) \) and \((-2,0)\).
- https://math.libretexts.org/Courses/Chabot_College/MTH_15%3A_Applied_Calculus_I/01%3A_Prerequisite_Review/1.05%3A_Quadratic_FunctionsFor the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[...For the horizontal intercepts, we solve for when the output will be zero: \[0=3x^2+5x-2.\nonumber \] In this case, the quadratic can be factored easily, providing the simplest method for solution.: \[0=(3x-1)(x+2),\nonumber \] so either \[ \begin{align*} 0 & = 3x-1\\ x & = \frac{1}{3} \end{align*} \nonumber \] or \[ \begin{align*} 0 & = x+2\\ x & = -2 \end{align*} \nonumber \] So the Horizontal intercepts are at \( \left(\frac{1}{3},0\right) \) and \((-2,0)\).
