Question

How do you write \[y=2{{x}^{2}}+3x-5\] in vertex form and identify the vertex, \[y\] intercept and \[x\] intercept?

Answer 1

Hint: From the question given, we had been given that \[y=2{{x}^{2}}+3x-5\]. And, we have been asked to find \[y=2{{x}^{2}}+3x-5\] in vertex form, and identify the vertex, \[y\] intercept and \[x\] intercept.

Complete step by step answer:
Now considering from the question we have \[y=2{{x}^{2}}+3x-5\]
First of all, to solve this we have to complete the square.
\[y=2{{x}^{2}}+3x-5\]
Shift \[-5\] to the left-hand side of the equation.
\[y+5=2{{x}^{2}}+3x\]
We know that vertex form is: \[y=a{{\left( x-h \right)}^{2}}+k\]where \[\left( -h,k \right)\] is the vertex.
We have to complete the square.
To complete the square, we need a leading coefficient of \[1\] and so factor out the \[2\].
By factoring out \[2\], we get \[y+5+c=2\left( {{x}^{2}}+\dfrac{3}{2}x+c \right)\]
Now \[c={{\left( \dfrac{1}{2}.\dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}}=\dfrac{9}{16}\]
\[\Rightarrow c=\dfrac{9}{16}\]
Remember that the \[c\] value must be multiplied by the \[2\] we factored out before we add it to the left side.
Now \[y+5+\left( 2.\dfrac{9}{16} \right)=2\left( {{x}^{2}}+\dfrac{3}{2}x+\dfrac{9}{16} \right)\]
\[\Rightarrow y+\dfrac{49}{8}=2{{\left( x+\dfrac{3}{4} \right)}^{2}}\]
Now, isolate the \[y\] and it is in the vertex form:
\[y=2{{\left( x+\dfrac{3}{4} \right)}^{2}}-\dfrac{49}{8}\]
We have been already discussed above that the vertex is \[\left( -h,k \right)\].
Therefore vertex \[=\left( -h,k \right)=\left( \dfrac{-3}{4},\dfrac{-49}{8} \right)\]
We know that \[y\] intercept is \[c\] for the standard form.
We know that the standard form is \[a{{x}^{2}}+bx+c\]
\[2{{x}^{2}}+3x-5\]
Therefore \[y\] intercept \[=-5\]
You must factor it to find the roots (\[x\] intercepts).
\[y=2{{x}^{2}}+3x-5\]
By the method of factorization, we get the above equation as,
\[\left( x-1 \right)\left( 2x+5 \right)\]
\[\Rightarrow x=1,-\dfrac{5}{2}\]
So, the \[x\] intercepts are \[1,-\dfrac{5}{2}\]
So, we got,
Vertex form: \[y=2{{\left( x+\dfrac{3}{4} \right)}^{2}}-\dfrac{49}{8}\]
Vertex \[=\left( -h,k \right)=\left( \dfrac{-3}{4},\dfrac{-49}{8} \right)\]
\[y\] intercept \[=-5\]
The \[x\] intercepts are \[1,-\dfrac{5}{2}\].
Therefore we can conclude that the $ x $ and $ y $ intercepts are $ \left( \dfrac{-5}{2},0 \right) $ or $ \left( 1,0 \right) $ and $ \left( 0,-5 \right) $ respectively.

Note:
We should be well aware of vertex form and intercepts. Also, we should be well known about converting the given equation into vertex form. Also, we should be well known about the process of finding intercepts. Also, we should be very careful while doing the calculation. This can be done in another way also but that is not preferred here because we have been asked to not do so. The other way is given by substituting $ x=0 $ gives the $ y $ intercept and substituting $ y=0 $ gives the $ x $ intercept. That is $ 2{{x}^{2}}+3x-5=y\Rightarrow 2\left( 0 \right)+3\left( 0 \right)-5=y\Rightarrow y=-5 $ so we can say $ y $ intercept is $ \left( 0,-5 \right) $ and $ 2{{x}^{2}}+3x-5=y\Rightarrow 0=2{{x}^{2}}+3x-5\Rightarrow x=\dfrac{-5}{2},1 $ so we can also say that $ x $ intercept $ \left( 1,0 \right) $ and $ \left( \dfrac{-5}{2},0 \right) $