How do you find the vertex and intercepts for $y = x^{2} - 4 x + 9$ $y = x^2 – 4x + 9$ ?

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How do you find the vertex and intercepts for $y = x^{2} - 4 x + 9$ $y = x^2 – 4x + 9$ ?

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Answer 1 · 2018-06-15T05:15:43

vertex: (2, 5)

x-intercept: none

y-intercept: (0, 9)

Explanation:

$y = x^{2} - 4 x + 9$ $y = x^2 - 4x + 9$

The equation is a quadratic equation in standard form, or $y = a x^{2} + b x + c$ $y = color(red)(a)x^2 + color(green)(b)x + color(blue)(c)$ .

The vertex is the minimum or maximum point of a parabola . To find the $x$ $x$ value of the vertex, we use the formula $x_{v} = - \frac{b}{2 a}$ $x_v = -color(green)(b)/(2color(red)(a))$ , where $x_{v}$ $x_v$ is the x-value of the vertex.

We know that $a = 1$ $color(red)(a = 1)$ and $b = - 4$ $color(green)(b = -4)$ , so we can plug them into the formula:
$x_{v} = \frac{- (- 4)}{2 (1)} = \frac{4}{2} = 2$ $x_v = (-(-4))/(2(1)) = 4/2 = 2$

To find the $y$ $y$ -value, we just plug in the $x$ $x$ value back into the equation:
$y = {(2)}^{2} - 4 (2) + 9$ $y = (2)^2 - 4(2) + 9$

Simplify:
$y = 4 - 8 + 9$ $y = 4 - 8 + 9$

$y = - 4 + 9$ $y = -4+9$

$y = 5$ $y = 5$

Therefore, the vertex is at $(2, 5)$ $(2, 5)$

$- - - - - - - - - - - - - - - - - - - -$ $--------------------$

Now to find the intercepts.

The $x$ $x$ -intercept is the value of $x$ $x$ when $y$ $y$ equals to zero.

To find it, just plug in $0$ $0$ for $y$ $y$ in the equation:
$0 = x^{2} - 4 x + 9$ $0 = x^2 - 4x + 9$

Since we cannot factor $x^{2} - 4 x + 9$ $x^2 - 4x + 9$ , we will use the quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-color(green)(b) +- sqrt(color(green)(b)^2-4color(red)(a)color(blue)(c))) / (2color(red)(a))$

We already know that $a = 1$ $color(red)(a = 1)$ , $b = - 4$ $color(green)(b = -4)$ , and $c = 9$ $color(blue)(c = 9)$ , so let's plug them into the formula:
$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (9)}}{2 (1)}$ $x = (-(color(green)(-4)) +- sqrt(color(green)((-4))^2 - 4(color(red)(1))(color(blue)(9))))/(2(color(red)(1))$

Now simplify:
$x = \frac{4 \pm \sqrt{16 - 36}}{2}$ $x = (4 +- sqrt(16 - 36))/2$

$x = \frac{4 \pm \sqrt{- 20}}{2}$ $x = (4+-sqrt(-20))/2$

Since we cannot do a square root of a negative number (it becomes imaginary), that means there are no $x$ $x$ -intercepts .

The $y$ $y$ -intercept is the value of $y$ $y$ when $x$ $x$ equals to zero.

To find it, just plug in $0$ $0$ for all the $x$ $x$ 's in the equation:
$y = {(0)}^{2} - 4 (0) + 9$ $y = (0)^2 - 4(0) + 9$

Simplify:
$y = 0 - 0 + 9$ $y = 0 - 0 + 9$

$y = 9$ $y = 9$

Now we write it as a coordinate, so it becomes $(0, 9)$ $(0, 9)$ .

In summary,

vertex: (2, 5)

x-intercept: none

y-intercept: (0, 9)

$- - - - - - - - - - - - - - - - - - - -$ $--------------------$

Here is a graph of this quadratic equation:
enter image source here

(desmos.com)

As you can see, there is no $x$ $x$ -intercept, and the vertex and $y$ $y$ -intercept are shown there.

For another explanation/example of finding the vertex and intercepts of a standard equation, feel free to watch this video:

Hope this helps!

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How do you find the vertex and intercepts for $y = x^{2} - 4 x + 9$ $y = x^2 – 4x + 9$ ?

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