# How do you write a quadratic function in standard form whose graph passes through points (1/2, -3/16), (3,-17/4), (4,-17/2)?

Feb 22, 2018

$f \left(x\right) = - \frac{3}{4} {x}^{2} + x - \frac{1}{2}$

#### Explanation:

Note that we can define a quadratic that takes the value $1$ when $x = \frac{1}{2}$ and the value $0$ when $x = 3$ or $x = 4$ by writing:

$\frac{\left(x - 3\right) \left(x - 4\right)}{\left(\frac{1}{2} - 3\right) \left(\frac{1}{2} - 4\right)} = \frac{\left(x - 3\right) \left(x - 4\right)}{\left(- \frac{5}{2}\right) \left(- \frac{7}{2}\right)} = \frac{4}{35} \left(x - 3\right) \left(x - 4\right)$

Similarly, we can define a quadratic that takes the value $1$ when $x = 3$ and the value $0$ when $x = \frac{1}{2}$ or $x = 4$ by writing:

$\frac{\left(2 x - 1\right) \left(x - 4\right)}{\left(6 - 1\right) \left(3 - 4\right)} = - \frac{1}{5} \left(2 x - 1\right) \left(x - 4\right)$

Similarly, we can define a quadratic that takes the value $1$ when $x = 4$ and the value $0$ when $x = \frac{1}{2}$ or $x = 3$ by writing:

$\frac{\left(2 x - 1\right) \left(x - 3\right)}{\left(8 - 1\right) \left(4 - 3\right)} = \frac{1}{7} \left(2 x - 1\right) \left(x - 3\right)$

We can add suitable multiples of these quadratics to get one with the desired properties:

$f \left(x\right) = \frac{4}{35} \left(x - 3\right) \left(x - 4\right) \left(- \frac{3}{16}\right) - \frac{1}{5} \left(2 x - 1\right) \left(x - 4\right) \left(- \frac{17}{4}\right) + \frac{1}{7} \left(2 x - 1\right) \left(x - 3\right) \left(- \frac{17}{2}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - \frac{3}{140} \left(x - 3\right) \left(x - 4\right) + \frac{17}{20} \left(2 x - 1\right) \left(x - 4\right) - \frac{17}{14} \left(2 x - 1\right) \left(x - 3\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{140} \left(- 3 \left(x - 3\right) \left(x - 4\right) + 119 \left(2 x - 1\right) \left(x - 4\right) - 170 \left(2 x - 1\right) \left(x - 3\right)\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{140} \left(- 3 \left({x}^{2} - 7 x + 12\right) + 119 \left(2 {x}^{2} - 9 x + 4\right) - 170 \left(2 {x}^{2} - 7 x + 3\right)\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{140} \left(- 3 {x}^{2} + 21 x - 36 + 238 {x}^{2} - 1071 x + 476 - 340 {x}^{2} + 1190 x - 510\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{140} \left(- 105 {x}^{2} + 140 x - 70\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{4} \left(- 3 {x}^{2} + 4 x - 2\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - \frac{3}{4} {x}^{2} + x - \frac{1}{2}$