Write the equation of a quadratic function with roots at 6 and 2, and a vertex at (4, −43). Write the equation in standard form. a) y=23x2−23x+6

Step 1 Root of the quadratic function =(6, 2) vertex of the quadratic function =(4, −43) Since we have given vertex the equation can be written as 1) y=a((x−4)2−43 Now put x=6 and y=0 in equation (1) we get that 0=a(6−4)2−430=4a−43 4a=43⇒a=13 Now put this value of a in equation (1) we get that y=13(x−4)2−43=13(x2+16−8x)−43 =13x2+163−8x3−43 =13x2+163−43−8x3 =13x2+4−8x3 y=13x2−8x3+4 Step 2 Therefore this is the required equation of quadratic function.

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High School
Algebra
Algebra I
Quadratic function and equation

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Answered question

2021-12-03

Write the equation of a quadratic function with roots at 6 and 2, and a vertex at

(4, \frac{- 4}{3})

. Write the equation in standard form.
a)

y = \frac{2}{3} x^{2} - \frac{2}{3} x + 6

Answer & Explanation

navratitehf

Beginner2021-12-04Added 20 answers

Step 1
Root of the quadratic function

= (6, 2)

vertex of the quadratic function

= (4, \frac{- 4}{3})

Since we have given vertex the equation can be written as
1)

y = a ({(x - 4)}^{2} - \frac{4}{3}

Now put

x = 6

and

y = 0

in equation (1) we get that

0 = a {(6 - 4)}^{2} - \frac{4}{3} 0 = 4 a - \frac{4}{3}

4 a = \frac{4}{3} \Rightarrow a = \frac{1}{3}

Now put this value of a in equation (1) we get that

y = \frac{1}{3} {(x - 4)}^{2} - \frac{4}{3} = \frac{1}{3} (x^{2} + 16 - 8 x) - \frac{4}{3}

= \frac{1}{3} x^{2} + \frac{16}{3} - \frac{8 x}{3} - \frac{4}{3}

= \frac{1}{3} x^{2} + \frac{16}{3} - \frac{4}{3} - \frac{8 x}{3}

= \frac{1}{3} x^{2} + 4 - \frac{8 x}{3}

y = \frac{1}{3} x^{2} - \frac{8 x}{3} + 4

Step 2
Therefore this is the required equation of quadratic function.

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