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Q.6

Expert-verifiedFound in: Page 566

Book edition
Student Edition

Author(s)
Carter, Cuevas, Day, Holiday, Luchin

Pages
801 pages

ISBN
9780078884801

**Consider $y={x}^{2}-5x+4$**

**Find the coordinates of the vertex. Is it a maximum or minimum point?**

The coordinate of the vertex is $\left(\frac{5}{2},\text{\hspace{0.17em}}-\frac{9}{4}\right)$.

A quadratic function, which is written in the form, $y=a{x}^{2}+bx+c$, where, $a\ne 0$ is called the standard form of the quadratic function

The graph of the function $y=a{x}^{2}+bx+c,$

Opens upward and has a minimum value at $x=-\frac{b}{2a}$, when $a>0$.

Opens downward and has a maximum value at $x=-\frac{b}{2a}$, when $a<0$.

The maximum or minimum point of the function $y=a{x}^{2}+bx+c$, is called the vertex.

Compare the quadratic function $y={x}^{2}-5x+4$ with the standard equation of the quadratic function, $y=a{x}^{2}+bx+c$.

$a=1,\text{\hspace{0.17em}}b=-5,\text{\hspace{0.17em}}c=4$

Substitute, $a=1$ and role="math" localid="1647752801168" $b=-5$ in $x=-\frac{b}{2a}$.

$x=-\frac{\left(-5\right)}{2\left(1\right)}\phantom{\rule{0ex}{0ex}}x=\frac{5}{2}$

Since,

So, the graph of the function opens upward and has a minimum point at $x=\frac{5}{2}$.

Substitute $x=\frac{5}{2}$ in $y={x}^{2}-5x+4$.

$y={\left(\frac{5}{2}\right)}^{2}-5\left(\frac{5}{2}\right)+4$

$=\frac{25}{4}-\frac{25}{2}+4\phantom{\rule{0ex}{0ex}}=\frac{25-2\left(25\right)+4\left(4\right)}{4}\phantom{\rule{0ex}{0ex}}=\frac{25-50+16}{4}\phantom{\rule{0ex}{0ex}}=\frac{41-50}{4}\phantom{\rule{0ex}{0ex}}=-\frac{9}{4}$

Hence, vertex $=\left(\frac{5}{2},\text{\hspace{0.17em}}-\frac{9}{4}\right)$

Therefore, the coordinate of the vertex is $\left(\frac{5}{2},\text{\hspace{0.17em}}-\frac{9}{4}\right)$.

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