Finding the x-intercepts (roots) of a parabola

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# Finding the x-intercepts (roots) of a parabola - PowerPoint PPT Presentation

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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## PowerPoint Slideshow about 'Finding the x-intercepts (roots) of a parabola' - edward-haney

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Presentation Transcript

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.

The number of real solutions for x can be:

No solutions

The discriminant will be negative

The number of real solutions for x can be:

One solution

The discriminant will be zero

The number of real solutions for x can be:

Two solutions

The discriminant will be positive

Identifying Solutions

Example f(x) = x2 - 4

0= x2 - 4

0= (x– 2)(x +2)

x = 2 or x = -2

Identifying Solutions

Now you try this problem.

f(x) = 2x - x2

0= 2x - x2

0 = x(2 – x)

x = 0 or x = 2

Identifying Solutions

Example f(x) = x2- 7x- 4

0= x2- 7x - 4

Cannot factor

Identifying Solutions

Example f(x) = x2 - 10x +25

0= x2- 10x + 25

0= (x – 5)(x – 5)

x = 5

Identifying Solutions

Example f(x) = x2 - 2x +60

0= x2 - 2x +60

Cannot factor

No x -intercepts