Find the Vertex f(t)=-(t-2)(t-15)

$f (t) = - (t - 2) (t - 15)$

Step 1

Rewrite the equation in terms of $x$ and $y$ .

$y = - (x - 2) (x - 15)$

Step 2

Rewrite the equation in vertex form.

Complete the square for $- (x - 2) (x - 15)$ .

Simplify the expression.

Apply the distributive property.

$(- x - - 2) (x - 15)$

Multiply $- 1$ by $- 2$ .

$(- x + 2) (x - 15)$

Expand $(- x + 2) (x - 15)$ using the FOIL Method.

Apply the distributive property.

$- x (x - 15) + 2 (x - 15)$

Apply the distributive property.

$- x \cdot x - x \cdot - 15 + 2 (x - 15)$

Apply the distributive property.

$- x \cdot x - x \cdot - 15 + 2 x + 2 \cdot - 15$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- (x \cdot x) - x \cdot - 15 + 2 x + 2 \cdot - 15$

Multiply $x$ by $x$ .

$- x^{2} - x \cdot - 15 + 2 x + 2 \cdot - 15$

Multiply $- 15$ by $- 1$ .

$- x^{2} + 15 x + 2 x + 2 \cdot - 15$

Multiply $2$ by $- 15$ .

$- x^{2} + 15 x + 2 x - 30$

Add $15 x$ and $2 x$ .

$- x^{2} + 17 x - 30$

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = 17$

$c = - 30$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{17}{2 \cdot - 1}$

Simplify the right side.

Multiply $2$ by $- 1$ .

$d = \frac{17}{- 2}$

Move the negative in front of the fraction.

$d = - \frac{17}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = - 30 - \frac{17^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $17$ to the power of $2$ .

$e = - 30 - \frac{289}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = - 30 - \frac{289}{- 4}$

Move the negative in front of the fraction.

$e = - 30 - - \frac{289}{4}$

Multiply $- - \frac{289}{4}$ .

Multiply $- 1$ by $- 1$ .

$e = - 30 + 1 (\frac{289}{4})$

Multiply $\frac{289}{4}$ by $1$ .

$e = - 30 + \frac{289}{4}$

To write $- 30$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = - 30 \cdot \frac{4}{4} + \frac{289}{4}$

Combine $- 30$ and $\frac{4}{4}$ .

$e = \frac{- 30 \cdot 4}{4} + \frac{289}{4}$

Combine the numerators over the common denominator.

$e = \frac{- 30 \cdot 4 + 289}{4}$

Simplify the numerator.

Multiply $- 30$ by $4$ .

$e = \frac{- 120 + 289}{4}$

Add $- 120$ and $289$ .

$e = \frac{169}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(x - \frac{17}{2})}^{2} + \frac{169}{4}$ .

$- {(x - \frac{17}{2})}^{2} + \frac{169}{4}$

Set $y$ equal to the new right side.

$y = - {(x - \frac{17}{2})}^{2} + \frac{169}{4}$

Step 3

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = - 1$

$h = \frac{17}{2}$

$k = \frac{169}{4}$

Step 4

Find the vertex $(h, k)$ .

$(\frac{17}{2}, \frac{169}{4})$

Step 5

Find the Vertex f(t)=-(t-2)(t-15)

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