How do I find the vertex form of $y= 5x^2-3x+2$

Question

I can't figure this out, I'm supposed to divide the second one by two but it turns into a fraction and then when I add it inside I'm supposed to find a number that when multiplied by itself gives 3/2 and when added gives a 3. How? https://youtu.be/pwITxyUghV0?t=378

Answer 1

Generally, speaking, this is the general plan you should have. $$y = ax^2+bx+c \implies y = a(x-h)^2+k \text{ when $(h, k)$ is the vertex }$$ All you need to do is find the vertex. The first step is to find its $x$-coordinate, which shown as $h$ in vertex form. $$h = -\frac{b}{2a}$$

For the sake of keeping everything clear, you can reach that formula by expanding the vertex form. $$a(x-h)^2+k \implies a(x^2-2xh+h^2)+k$$ $$\implies ax^2-2axh+h^2a+k \implies ax^2-2axh+(h^2a+k)$$

Notice how we’ve expanded it into $ax^2+bx+c$ form. Now, comparing $bx$ and $-2axh$, we reach the formula. $$bx = -2axh \implies b = -2ah \implies h = -\frac{b}{2a}$$

Here, the function is $y = 5x^2-3x+2$. To find $h$, use the formula. $$h = -\frac{-3}{2(5)} = \frac{3}{10}$$

Finding $k$ is just a matter of plugging in $h$ in the function and solving for $y$.

$$y = 5\biggr(\frac{3}{10}\biggr)^2-3\biggr(\frac{3}{10}\biggr)+2 \implies y = \frac{31}{20} \text{ ,so } k = \frac{31}{20}$$

Now, just enter the values.

$$y = a(x-h)^2+k \implies \boxed{y = 5\biggr(x-\frac{3}{10}\biggr)^2+\frac{31}{20}}$$

Answer 2

The vertex form is $$y=a(x-h)^2 + k$$ which expands to $$ax^2-(2ah)x+(ah^2+k)$$ Thus, our goal is to find the values of $a,h,k$ for which the relation $$ax^2-(2ah)x+(ah^2+k)=5x^2-3x+2$$ is an identity.

Equating leading coefficients, we get $a=5$.

Then, since $a=5$, equating the coefficients of $x$ yields $h={\large{\frac{3}{10}}}$.

Then, since $a=5$ and $h={\large{\frac{3}{10}}}$, equating the constant terms yields $k={\large{\frac{31}{20}}}$.

Hence the vertex form is $$y=5\left(x-{\small{\frac{3}{10}}}\right)^{\!2}+{\small{\frac{31}{20}}}$$

To automate the process for future calculations, if we apply the same procedure to the equation $$y=ax^2+bx+c$$ where $a,b,c$ are regarded as unknown constants, with $a\ne 0$, we get the vertex form $$y=a(x-h)^2+k$$ where $a$ is the same for both, and $h,k$ are given by \begin{align*} h&=-\frac{b}{2a}\\[6pt] k&=c-\frac{b^2}{4a}\\[4pt] \end{align*} Typically, we only memorize the formula for $h$, and then, letting $$f(x)=ax^2+bx+c$$ we solve for $k$ by the formula $k=f(h)$, which can be written as $$k=f\bigl(-{\small{\frac{b}{2a}}}\bigr)$$ Applying these formulas to the given equation $$y=5x^2-3x+2$$ we get \begin{align*} a&=5\\[4pt] h&=-{\small{\frac{b}{2a}}}={\small{\frac{3}{10}}}\\[4pt] k&=f\bigl(-{\small{\frac{b}{2a}}}\bigr) =f\bigl({\small{\frac{3}{10}}}\bigr) ={\small{\frac{31}{20}}}\\[4pt] \end{align*}

Answer 3

Let $x:=t+a$ to translate the curve horizontally by the amount $a$.

$$y=5x^2-3x+2=5(t+a)^2-3(t+a)+2=5t^2+(10a-3)t+5a^2-3a+2.$$

Now if you let the linear term vanish,

$$10a-3=0,$$

the equation turns to

$$y=5t^2+5\frac{9}{100}-3\frac3{10}+2=5t^2+\frac{31}{20},$$

which is

$$y=5\left(x-\frac3{10}\right)^2+\frac{31}{20}.$$

Answer 4

To give an alternate look at the same process as the other answers, if you understand completing the square, then you can just use that to guide your work. The trick is to either "ignore" or "move" the extra stuff out of the way before doing the regular process.

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Ignoring it looks like this: $$ 5x^2-3x+2 = 5(x^2 - \frac{3}{5}x) + 2 $$ then perform the "completing the square" operation on $x^2 - \frac{3}{5}x$ to get $$ 5\left(\left(x-\frac{3}{10}\right)^2 - \frac{9}{100}\right) + 2 = 5\left(x-\frac{3}{10}\right)^2 + \frac{31}{20} $$

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Moving it looks like this: $$ y = 5x^2-3x+2 \Rightarrow \frac{1}{5}y - \frac{2}{5} = x^2-\frac{3}{5}x $$ then perform the "completing the square" operation on $x^2-\frac{3}{5}x$ to get $$ \frac{1}{5}y - \frac{2}{5} + \frac{9}{100} = x^2 - \frac{3}{5}x + \frac{9}{100} = \left(x-\frac{3}{10}\right)^2 \Rightarrow y = 5\left(x-\frac{3}{10}\right)^2 + \frac{31}{20} $$

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In both cases, the idea is to isolate the $x^2-\frac{b}{a}x$ term, then perform the "completing the square" operation on that term, before moving everything around the new square term to get the vertex form.

Answer 5

To find the vertex form of any quadratic equation, one must complete the square for that equation. If one is given an equation in vertex form, $a(x-h)^2+k$ for example, the vertex will be $(h,k)$. Here is an answer which tells you how to complete the square.

But here's another way to find the vertex without having to complete the square.

Suppose you are given a parabola $f(x)=ax^2+bx+c$. From calculus, we know that the $x$-value of the vertex of the parabola is the point at which $f'(x)=2ax+b$, the derivative of $f(x)$, crosses the $x$-axis. To find this point, we must set $f'(x)=0$, and solve for $x$. $$f'(x)=0$$ $$2ax+b=0$$ $$2ax=-b$$ $$x=\frac{-b}{2a}$$ The, we know that the $y$-value of the vertex will be given by $f(\frac{-b}{2a})$. Hence, $$f(\frac{-b}{2a})$$ $$=a(\frac{-b}{2a})^2+b(\frac{-b}{2a})+c$$ $$=\frac{b^2 a}{4a^2}-\frac{b^2}{2a}+c$$ $$=\frac{b^2}{4a}-\frac{b^2}{2a}+c$$ $$=\frac{b^2}{2a}(\frac{1}{2}-1)+c$$ $$=c-\frac{b^2}{4a}$$ Thus, the vertex of $ax^2+bx+c$ is given by $$(\frac{-b}{2a},c-\frac{b^2}{4a})$$