Find the equation of a quadratic function with zeros at x = 2 and x = 4 and y- intercept (0,8). Give your answer in standard form.

Since we know the zeroes of the function we can write:

y = a(x - 2)(x - 4) but we need to solve for a. Thankfully, we know that (0, 8) belongs to the line so we can plug these values in for x and y to solve for a. We get:

8 = a(0 - 2)(0 - 4)

8 = a * (-2) * (-4)

8 = 8a

a = 1 so the equation is y = 1(x - 2)(x - 4) or just y = (x - 2)(x - 4) for short.

Turning this into standard form we get y = x² - 6x + 8.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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