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Question: How do you convert from vertex form to intercept form of \(y-4=-{{\left( x-2 \right)}^{2}}\)?...

How do you convert from vertex form to intercept form of y−4=−(x−2)2y-4=-{{\left( x-2 \right)}^{2}}?

Explanation

Solution

We use the fact that the vertex form of quadratic equation y=ax2+bx+cy=a{{x}^{2}}+bx+c is given by (x−h)2=4a(y−k){{\left( x-h \right)}^{2}}=4a\left( y-k \right) where (h,k)\left( h,k \right) is the vertex of the parabola and the intercept form of the quadratic equation is given by y=k(x−α)(x−β)y=k\left( x-\alpha \right)\left( x-\beta \right) where α,β\alpha ,\beta are the x−x-intercepts of the parabola. We find the roots of the given quadratic equation to find the intercept form. $$$$

Complete step-by-step answer:
We know that the general quadratic equation is given by ax2+bx+c=0a{{x}^{2}}+bx+c=0 where graph of the quadratic polynomial y=ax2+bx+cy=a{{x}^{2}}+bx+c represents a parabola in xy−xy-plane. The vertex from of the parabola is given by with vertex (h,k)\left( h,k \right) is given by
(x−h)2=4a(y−k){{\left( x-h \right)}^{2}}=4a\left( y-k \right)
We are given the following quadratic equation in the question
y−4=−(x−2)2y-4=-{{\left( x-2 \right)}^{2}}
We expand the right hand side using the algebraic identity (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}to have

& \Rightarrow y-4=-\left( {{x}^{2}}-8x+16 \right) \\\ & \Rightarrow y=-{{x}^{2}}+8x-12 \\\ \end{aligned}$$ So we need to convert the above result to the intercept form $y=k\left( x-\alpha \right)\left( x-\beta \right)$. So we have to find the roots since we have $x-$intercepts at $y=0$. So let us have $$\begin{aligned} & y=0 \\\ & \Rightarrow -{{x}^{2}}+8x-12=0 \\\ & \Rightarrow {{x}^{2}}-8x-12=0 \\\ \end{aligned}$$ We use the splitting the middle term method to have; $$\begin{aligned} & \Rightarrow {{x}^{2}}-6x-2x+12=0 \\\ & \Rightarrow x\left( x-6 \right)-2\left( x-6 \right)=0 \\\ & \Rightarrow \left( x-6 \right)\left( x-2 \right)=0 \\\ & \Rightarrow x=6,2 \\\ \end{aligned}$$ So we have the roots of the quadratic polynomial as $\alpha =6,\beta =2$. Since we have multiplied $-1$ both sides and then solved the equation we have the required intercept form is $$y=-\left( x-2 \right)\left( x-6 \right)$$ **Note:** We note that the vertex of the given equation $y-4=-{{\left( x-2 \right)}^{2}}$ is at $\left( 2,4 \right)$ and also note that at the vertex the minimum values of the quadratic polynomial occurs. So at $x=4$ we are getting the minimum as $y=-\left( 4-2 \right)\left( 4-6 \right)=4$. So our equation in intercept form is verified. We can find the vertex from general quadratic polynomial $y=a{{x}^{2}}+bx+c$ at $x=\dfrac{-b}{2a}$ . If in $y=a{{x}^{2}}+bx+c$ we have $a > 0$ we get a parabola opens upwards and if we have $a < 0$ we get a parabola opens downwards.