Question Rewrite a quadratic expression in standard, vertex, and factored forms. Evaluate the expression at x=0 and x=2.
(a) $a x^{2} + b x + c$ (b) $k(a x + b)(c x + d)$ (c) Evaluate $\boxed{(x-2)^{2}-25}$ at $x=0$ (d) Evaluate $\boxed{(x-2)^{2}-25}$ at $x=2$

Studdy Solution

STEP 1

Assumptions1. The quadratic expression is given in vertex form as $(x-)^{}-25$ . We are to write the expression in standard form $ax^{}+bx+c$
3. We are to write the expression in factored form $k(ax+b)(cx+d)$ 4. We are to evaluate the expression at $x=0$ and $x=$ using each of the three forms and compare the results

STEP 2

To write the expression in standard form, we need to expand the square in the vertex form.
$(x-2)^{2}-25 = x^{2} -4x +4 -25$

STEP 3

implify the expression to get the standard form.
$x^{2} -x + -25 = x^{2} -x -21$ So, the standard form of the expression is $x^{2} -x -21$ .

STEP 4

To write the expression in factored form, we need to factorize the standard form.
The factored form of a quadratic expression $ax^{2}+bx+c$ is $a(x-p)(x-q)$ , where $p$ and $q$ are the roots of the equation $ax^{2}+bx+c=0$ .

STEP 5

To find the roots of the equation, we can use the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ .
Plugging in $a=1$ , $b=-4$ , and $c=-21$ into the quadratic formula, we get $x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-21)}}{2(1)}$

STEP 6

implify the expression to get the roots.
$x = \frac{4 \pm \sqrt{16+84}}{2}$

STEP 7

Further simplify the expression to get the roots.
$x = \frac{4 \pm \sqrt{100}}{2}$

STEP 8

Calculate the roots.
$x = \frac{4 \pm10}{2}$ So, the roots are $x=7$ and $x=-3$ .

STEP 9

Now, we can write the expression in factored form as $k(x-p)(x-q)$ .
$(x-2)^{2}-25 = (x-7)(x+3)$

STEP 10

To evaluate the expression at $x=0$ , we substitute $x=0$ into each of the three forms and calculate the result.
In vertex form $(0-2)^{2}-25 =4 -25 = -21$ In standard form $0^{2} -4*0 -21 = -21$ In factored form $(0-7)(0+3) = -21$ So, the expression evaluates to $-21$ at $x=0$ in all three forms.

STEP 11

To evaluate the expression at $x=$ , we substitute $x=$ into each of the three forms and calculate the result.
In vertex form $(-)^{}-25 =0 -25 = -25$ In standard form $^{} -4* -21 =4 -8 -21 = -25$ In factored form $(-7)(+3) = -5*5 = -25$ So, the expression evaluates to $-25$ at $x=$ in all three forms.

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Question Rewrite a quadratic expression in standard, vertex, and factored forms. Evaluate the expression at x=0 and x=2.
(a) $a x^{2} + b x + c$ (b) $k(a x + b)(c x + d)$ (c) Evaluate $\boxed{(x-2)^{2}-25}$ at $x=0$ (d) Evaluate $\boxed{(x-2)^{2}-25}$ at $x=2$

Studdy Solution

STEP 1

STEP 2

To write the expression in standard form, we need to expand the square in the vertex form.
$(x-2)^{2}-25 = x^{2} -4x +4 -25$

STEP 3

implify the expression to get the standard form.
$x^{2} -x + -25 = x^{2} -x -21$ So, the standard form of the expression is $x^{2} -x -21$ .

STEP 4

To write the expression in factored form, we need to factorize the standard form.
The factored form of a quadratic expression $ax^{2}+bx+c$ is $a(x-p)(x-q)$ , where $p$ and $q$ are the roots of the equation $ax^{2}+bx+c=0$ .

STEP 5

To find the roots of the equation, we can use the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ .
Plugging in $a=1$ , $b=-4$ , and $c=-21$ into the quadratic formula, we get $x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-21)}}{2(1)}$

STEP 6

implify the expression to get the roots.
$x = \frac{4 \pm \sqrt{16+84}}{2}$

STEP 7

Further simplify the expression to get the roots.
$x = \frac{4 \pm \sqrt{100}}{2}$

STEP 8

Calculate the roots.
$x = \frac{4 \pm10}{2}$ So, the roots are $x=7$ and $x=-3$ .

STEP 9

Now, we can write the expression in factored form as $k(x-p)(x-q)$ .
$(x-2)^{2}-25 = (x-7)(x+3)$

STEP 10

To evaluate the expression at $x=0$ , we substitute $x=0$ into each of the three forms and calculate the result.
In vertex form $(0-2)^{2}-25 =4 -25 = -21$ In standard form $0^{2} -4*0 -21 = -21$ In factored form $(0-7)(0+3) = -21$ So, the expression evaluates to $-21$ at $x=0$ in all three forms.

STEP 11

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Studdy Solution

STEP 1

STEP 2

To write the expression in standard form, we need to expand the square in the vertex form.(x−2)2−25=x2−4x+4−25 (x-2)^{2}-25 = x^{2} -4x +4 -25 (x−2)2−25=x2−4x+4−25

STEP 3

implify the expression to get the standard form.x2−x+−25=x2−x−21 x^{2} -x + -25 = x^{2} -x -21 x2−x+−25=x2−x−21So, the standard form of the expression is x2−x−21x^{2} -x -21x2−x−21.

STEP 4

To write the expression in factored form, we need to factorize the standard form.The factored form of a quadratic expression ax2+bx+cax^{2}+bx+cax2+bx+c is a(x−p)(x−q)a(x-p)(x-q)a(x−p)(x−q), where ppp and qqq are the roots of the equation ax2+bx+c=0ax^{2}+bx+c=0ax2+bx+c=0.

STEP 5

STEP 6

implify the expression to get the roots.x=4±16+842x = \frac{4 \pm \sqrt{16+84}}{2}x=24±16+84​​

STEP 7

Further simplify the expression to get the roots.x=4±1002x = \frac{4 \pm \sqrt{100}}{2}x=24±100​​

STEP 8

Calculate the roots.x=4±102x = \frac{4 \pm10}{2}x=24±10​So, the roots are x=7x=7x=7 and x=−3x=-3x=−3.

STEP 9

Now, we can write the expression in factored form as k(x−p)(x−q)k(x-p)(x-q)k(x−p)(x−q).(x−2)2−25=(x−7)(x+3) (x-2)^{2}-25 = (x-7)(x+3) (x−2)2−25=(x−7)(x+3)

STEP 10

STEP 11

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To write the expression in standard form, we need to expand the square in the vertex form.
$(x-2)^{2}-25 = x^{2} -4x +4 -25$

implify the expression to get the standard form.
$x^{2} -x + -25 = x^{2} -x -21$ So, the standard form of the expression is $x^{2} -x -21$ .

To write the expression in factored form, we need to factorize the standard form.
The factored form of a quadratic expression $ax^{2}+bx+c$ is $a(x-p)(x-q)$ , where $p$ and $q$ are the roots of the equation $ax^{2}+bx+c=0$ .

implify the expression to get the roots.
$x = \frac{4 \pm \sqrt{16+84}}{2}$

Further simplify the expression to get the roots.
$x = \frac{4 \pm \sqrt{100}}{2}$

Calculate the roots.
$x = \frac{4 \pm10}{2}$ So, the roots are $x=7$ and $x=-3$ .

Now, we can write the expression in factored form as $k(x-p)(x-q)$ .
$(x-2)^{2}-25 = (x-7)(x+3)$