Match the following quadratic polynomials with the coordinates of their vertex.

(\frac{11}{14}, \frac{- 149}{28})

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(- 1, - 2)

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(\frac{1}{3}, \frac{8}{3})

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(\frac{2}{9}, \frac{- 41}{9})

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Solution

For any quadratic polynomial $y = a x^{2} + b x + c,$ the coordinates of the vertex is given as:
$(- \frac{b}{2 a}, - \frac{D}{4 a}) \dots (i)$
Now, let's take the quadratic expressions one by one to find it's vertex.
$(1) . y = 5 x^{2} + 10 x + 3$
On comparing with standard form of quadratic expression $y = a x^{2} + b x + c$
we get, $a = 5, b = 10, c = 3$
$D = b^{2} - 4 a c = (10)^{2} - 4 \cdot 5 \cdot 3 = 40$
Substituting the values of $a, b & D$ in $(i),$ we get:
$(\frac{- b}{2 a}, \frac{- D}{4 a}) \equiv (\frac{- 10}{2 \cdot 5}, \frac{- 40}{4 \cdot 5}) \equiv (- 1, - 2)$
Hence, the coordinates of vertex is given as:
$(- 1, - 2) .$

$(2) . y = 3 x^{2} - 2 x + 3$
On comparing with standard form of quadratic expression $y = a x^{2} + b x + c$
we get, $a = 3, b = - 2, c = 3$
$D = b^{2} - 4 a c = (- 2)^{2} - 4 \cdot 3 \cdot 3 = - 32$
Substituting the values of $a, b & c$ in $(i),$ we get:
$(\frac{- b}{2 a}, \frac{- D}{4 a}) \equiv (\frac{- (- 2)}{2 \cdot 3}, \frac{- (- 32)}{4 \cdot 3}) \equiv (\frac{1}{3}, \frac{8}{3})$
Hence, the coordinates of vertex is given as:
$(\frac{1}{3}, \frac{8}{3}) .$

$(3) . y = 7 x^{2} - 11 x - 1$
On comparing with standard form of quadratic expression $y = a x^{2} + b x + c$
we get, $a = 7, b = - 11, c = - 1$
$D = b^{2} - 4 a c = (- 11)^{2} - 4 \cdot 7 \cdot (- 1) = 149$
Substituting the values of $a, b & c$ in $(i),$ we get:
$(\frac{- b}{2 a}, \frac{- D}{4 a}) \equiv (\frac{- (- 11)}{2 \cdot 7}, \frac{- 149}{4 \cdot 7}) \equiv (\frac{11}{14}, \frac{- 149}{28})$
Hence, the coordinates of vertex is given as:
$(\frac{11}{14}, \frac{- 149}{28}) .$

$(4) . y = - 9 x^{2} + 4 x - 5$
On comparing with standard form of quadratic expression $y = a x^{2} + b x + c$
we get, $a = - 9, b = 4, c = - 5$
$D = b^{2} - 4 a c = (4)^{2} - 4 \cdot (- 9) \cdot (- 5) = - 164$
Substituting the values of $a, b & c$ in $(i),$ we get:
$(\frac{- b}{2 a}, \frac{- D}{4 a}) \equiv (\frac{- 4}{2 \cdot (- 9)}, \frac{- (- 164)}{4 \cdot (- 9)}) \equiv (\frac{2}{9}, \frac{- 41}{9})$
Hence, the coordinates of vertex is given as:
$(\frac{2}{9}, \frac{- 41}{9}) .$

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