Algebra 2 help . Use y = x2-8x+7?

y = x² - 8x + 7

Tha'ts a quadratic function in the form y = ax² + bx + c, where a = 1, b = -8, c = 7.

a. The discriminant is represented by Δ.

Δ = b² - 4ac = (-8)² - 4(1)(7) = 64 - 28 = 36

b. The discriminant tells you the type of roots. If Δ is negative, the equation has two complex root. If Δ is zero, the equation has one repeated root (also called a double root). If Δ is positive, the equation has two real and distinct roots. For real and distinct roots, if Δ is a perfect square (like 1, 4, 9, 16, etc) then the real roots are rational; otherwise they are irrational.

For this equation Δ = 36. So there are two real and rational roots.

c. The roots occur wher y = 0, so

y = 0 = x² - 8x + 7

(x - 1)(x - 7) = 0

x = 1, x = 7 are the roots

d. x = (-b ± √Δ)/2a = (8 ± √36)/2

x = (8 + 6)/2 = 14/2 = 7

x = (8 - 6)/2 = 2/2 = 1

f. The y-intercept occurs at x = 0

y = x² - 8x + 7 = (0)² - 8(0) + 7 = 7

g. y = x² - 8x + 7

y = (x² - 8x + ___ ) + ___+ 7

y = (x² - 8x + 16) - 16 + 7 = (x - 4)² - 9

h. Completing the square form is the same as vertex form. Change the sign for the number in parentheses to get the x-coordinate. The y-coordinate is the number outside the parentheses just as it is. The vertex is (4, -9).

b. it will have two real roots because just use the vertex -b/2a and if it goes below the x axis it will tell you whethere it has real or nonreal solutions

d..quadratic formula you'll end up with two real rational zeros of (1,0) and (7,0) as the x-intercepts

f. yint is just the (0,7) from the last number of the equation

h. -b/2a = 8/2 = 4> take that four and plug it into x > 16-32+7=-9 so the vertex is (4,-9)

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