# A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function. -The quadratic coefficients is -The linear coefficients is -The constant term is

A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function.
-The linear coefficients is
-The constant term is
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Standard vertex form of quadratic function is
$y=a{\left(x-h\right)}^{2}+k$
Where point (h,k) is vertex point.
$\because$ vertex point is (1,6)
$\because$ Substitute this point in standard vertex form
$y=a{\left(x-1\right)}^{2}+6$
$=a{\left(x-1\right)}^{2}+6\left(1\right)$
Also function pass through point (−2,−3).
$\because -3=a{\left(-2-1\right)}^{2}+6$
$⇒-3=a\cdot \left(-3\right)2+6$
$⇒-3=a\cdot 9+6$
$⇒9a=-3-6$
$⇒9a=-9$
$⇒a=-\frac{9}{9}$
=-1
Substitute 'a' value in equation (1)
$y=\left(-1\right)\cdot {\left(x-1\right)}^{2}+6$
$=-\left({x}^{2}+1-2x\right)+6$
$=-{x}^{2}+2x-1+6$
$={x}^{2}+2x+5$
Hence quadratic function is $y=-{x}^{2}$
Where
Linear coefficient=2
Constant term$={x}^{2}+2x+5$