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Question: How do you write \[y-5=-2{{\left( x+1 \right)}^{2}}\] in intercept form?...

How do you write y−5=−2(x+1)2y-5=-2{{\left( x+1 \right)}^{2}} in intercept form?

Explanation

Solution

Keep y as it is in the L.H.S. and take the constant term -5 to the R.H.S. Now, take -2 common from all the terms in the R.H.S. obtained and simplify the expression into the form a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to get the intercept form of the given quadratic expression.

Complete step-by-step solution:
Here, we have been provided with the expression: y−5=−2(x+1)2y-5=-2{{\left( x+1 \right)}^{2}} and we are asked to write it in intercept form. But first we need to know the general intercept form.
Now, here as we can see that the given expression is a quadratic equation in x. So, leaving y in the L.H.S. and taking the constant term -5 to the R.H.S., we get,
⇒y=−2(x+1)2+5\Rightarrow y=-2{{\left( x+1 \right)}^{2}}+5
So, the intercept form of a quadratic equation is given as: - a(x−m)(x−n)a\left( x-m \right)\left( x-n \right) where ‘a’ is any constant other than 0 and m and n are called x – intercepts. They are the points where the parabola represented by the quadratic expression cuts the x – axis. So, let us convert our expression into this form. So, we have,
∵y=−2(x+1)2+5\because y=-2{{\left( x+1 \right)}^{2}}+5
Taking -2 common from both the terms, we get,
⇒y=−2[(x+1)2−52]\Rightarrow y=-2\left[ {{\left( x+1 \right)}^{2}}-\dfrac{5}{2} \right]
We can write 52=(52)2\dfrac{5}{2}={{\left( \sqrt{\dfrac{5}{2}} \right)}^{2}}, so we get,
⇒y=−2[(x+1)2−(52)2]\Rightarrow y=-2\left[ {{\left( x+1 \right)}^{2}}-{{\left( \sqrt{\dfrac{5}{2}} \right)}^{2}} \right]
Using the algebraic identity: - a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right), we get,
⇒y=−2[(x+1+52)(x+1−52)]\Rightarrow y=-2\left[ \left( x+1+\sqrt{\dfrac{5}{2}} \right)\left( x+1-\sqrt{\dfrac{5}{2}} \right) \right]
⇒y=−2[(x+(1+52))(x+(1−52))]\Rightarrow y=-2\left[ \left( x+\left( 1+\sqrt{\dfrac{5}{2}} \right) \right)\left( x+\left( 1-\sqrt{\dfrac{5}{2}} \right) \right) \right]
On comparing the above obtained relation with y=a(x−m)(x−n)y=a\left( x-m \right)\left( x-n \right), we get,

& \Rightarrow a=-2 \\\ & \Rightarrow m=-\left( 1+\sqrt{\dfrac{5}{2}} \right) \\\ & \Rightarrow n=-\left( 1-\sqrt{\dfrac{5}{2}} \right) \\\ \end{aligned}$$ So, x – intercepts are $$-\left( 1+\sqrt{\dfrac{5}{2}} \right)$$ and $$-\left( 1-\sqrt{\dfrac{5}{2}} \right)$$ and these are the points where the graph of the equation $$y=-2{{\left( x+1 \right)}^{2}}+5$$ will cut the x – axis. **Hence, $$y=-2\left[ \left( x+\left( 1+\sqrt{\dfrac{5}{2}} \right) \right)\left( x+\left( 1-\sqrt{\dfrac{5}{2}} \right) \right) \right]$$ is the intercept form and our answer.** **Note:** One may apply another method to get the answer. What you can do is substitute the quadratic equation in x equal to zero and solve it using the discriminant formula. Assume the two values of x obtained as x = m and x = n and then consider the product $$\left( x-m \right)\left( x-n \right)$$ to get the intercept form. At the end multiply this product with -2 to balance the equation so that the original form does not change. Note that here we will have difficulty in applying the middle term split method because it will be difficult to think of the terms like $$-\left( 1+\sqrt{\dfrac{5}{2}} \right)$$ and $$-\left( 1-\sqrt{\dfrac{5}{2}} \right)$$ as the roots.