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Finding the x-intercepts (roots) of a parabola

Finding the x-intercepts (roots) of a parabola. Quadratic Equation. y = ax 2 + bx + c When we talk about finding the roots we want to find the the value of x when y = 0. So we substitute zero for y and solve the resulting equation. Quadratic Solutions.

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Finding the x-intercepts (roots) of a parabola

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  1. Finding the x-intercepts (roots)of a parabola

  2. Quadratic Equation y = ax2 + bx + c When we talk about finding the roots we want to find the the value of x when y = 0. So we substitute zero for y and solve the resulting equation.

  3. Quadratic Solutions The number of real solutions for x can be: No solutions The discriminant will be negative

  4. Quadratic Solutions The number of real solutions for x can be: One solution The discriminant will be zero

  5. Quadratic Solutions The number of real solutions for x can be: Two solutions The discriminant will be positive

  6. Identifying Solutions Example f(x) = x2 - 4 0= x2 - 4 0= (x– 2)(x +2) x = 2 or x = -2

  7. Identifying Solutions Now you try this problem. f(x) = 2x - x2 0= 2x - x2 0 = x(2 – x) x = 0 or x = 2

  8. Identifying Solutions Example f(x) = x2- 7x- 4 0= x2- 7x - 4 Cannot factor

  9. Identifying Solutions Example f(x) = x2 - 10x +25 0= x2- 10x + 25 0= (x – 5)(x – 5) x = 5

  10. Identifying Solutions Example f(x) = x2 - 2x +60 0= x2 - 2x +60 Cannot factor No x -intercepts

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