How do you write a quadratic function in standard form whose graph passes through points (1/2, -3/10), (1, 6/5), (1/4, -3/10)?

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How do you write a quadratic function in standard form whose graph passes through points (1/2, -3/10), (1, 6/5), (1/4, -3/10)?

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Answer 1 · 2017-04-18T01:29:22

Start with the vertex form and use symmetry to find the x coordinate.
Use the points to find the other 2 two values of the vertex form.
Expand the vertex form into the standard form.

Explanation:

Because you specified a function we discard the vertex form $x = a {(y - k)}^{2} + h$ $x = a(y-k)^2 + h$ and use only the form:

$y = a {(x - h)}^{2} + k [1]$ $y = a(x-h)^2+k" [1]"$

Because the we know the 2 x values corresponding to y value, $- \frac{3}{10}$ $-3/10$ we know know that the x coordinate of the vertex is halfway between $\frac{1}{4} and \frac{1}{2}$ $1/4 and 1/2$

$h = \frac{\frac{1}{2} + \frac{1}{4}}{2}$ $h = (1/2+1/4)/2$

$h = \frac{3}{8}$ $h = 3/8$

Substitute $\frac{3}{8}$ $3/8$ for h into equation [1]:

$y = a {(x - \frac{3}{8})}^{2} + k [2]$ $y = a(x-3/8)^2+k" [2]"$

Substitute the point $(\frac{1}{2}, - \frac{3}{10})$ $(1/2,-3/10)$ into equation [2]:

$- \frac{3}{10} = a {(\frac{1}{2} - \frac{3}{8})}^{2} + k$ $-3/10 = a(1/2-3/8)^2+k$

$- \frac{3}{10} = a {(\frac{1}{8})}^{2} + k$ $-3/10 = a(1/8)^2+k$

$- \frac{3}{10} = \frac{a}{64} + k [3]$ $-3/10 = a/64+k" [3]"$

Substitute the point $(1, \frac{6}{5})$ $(1,6/5)$ into equation [2]:

$\frac{6}{5} = a {(1 - \frac{3}{8})}^{2} + k$ $6/5 = a(1-3/8)^2+k$

$\frac{6}{5} = a {(\frac{5}{8})}^{2} + k$ $6/5 = a(5/8)^2+k$

$\frac{6}{5} = \frac{25 a}{64} + k [4]$ $6/5 = (25a)/64+k" [4]"$

Subtract equation [3] from equation [4]:

$\frac{6}{5} + \frac{3}{10} = \frac{24 a}{64}$ $6/5+3/10 = (24a)/64$

$\frac{15}{10} = \frac{24}{64} a$ $15/10 = 24/64a$

$a = \frac{3}{2} (\frac{64}{24})$ $a = 3/2(64/24)$

$a = \frac{1}{2} (\frac{64}{8})$ $a = 1/2(64/8)$

$a = 4$ $a = 4$

Use equation [3] to find the value of k:

$- \frac{3}{10} = \frac{4}{64} + k$ $-3/10 = 4/64+k$

$- \frac{3}{10} = \frac{1}{16} + k$ $-3/10 = 1/16+k$

$k = - \frac{3}{10} - \frac{1}{16}$ $k = -3/10 - 1/16$

$k = - \frac{29}{80}$ $k = -29/80$

Substitute the value of "a" and "k" into equation [2]:

$y = 4 {(x - \frac{3}{8})}^{2} - \frac{29}{80}$ $y = 4(x-3/8)^2-29/80$

Expand the square:

$y = 4 (x^{2} - \frac{6}{8} x + \frac{9}{64}) - \frac{29}{80}$ $y = 4(x^2 - 6/8x +9/64)-29/80$

Distribute the 4:

$y = 4 x^{2} - 3 x + \frac{9}{16} - \frac{29}{80}$ $y = 4x^2 - 3x +9/16-29/80$

This is the standard form:

$y = 4 x^{2} - 3 x + \frac{1}{5}$ $y = 4x^2 - 3x +1/5$

The following is a graph of the equation and the 3 points:

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How do you write a quadratic function in standard form whose graph passes through points (1/2, -3/10), (1, 6/5), (1/4, -3/10)?

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