Find the equation of the line tangent to the following curve at $x=1$. Write your answer in $y=mx+b$ format. The curve is defined by $y=2x^2+6x-4.$
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[Math] Find the equation of the line tangent to the following curve at x=1. Write your answer in y=mx+b format.
calculus
Best Answer
Step 1: Find the slope of the tangent to the curve $y=2x^2+6x−4.$.
$$m=\dfrac{dy}{dx}=\dfrac{d}{dx}(2x^2+6x−4)=4x+6.$$
At $x=1$ the slope is $m=4+6=10$.
Step 2: Write the equation of the tangent in the form $y=mx+c$.
$$y=10x+c \tag{1}$$ Step 4: Find $c$.
To find $c$ notice that the intersection point of the tangent with the curve is known. The intersection point has $x$ coordinate $1$ and y coordinate $y=2(1)^2+6\cdot 1-4=4$. So we know that the point $(1,4)$ lie on the tangent.
We can find $c$ now. For this substitute $x=1$ and $y=4$ in the eqation 1.
$$4=10(1)+c \implies \ c=-6$$.
Step 5: Write down the answer, that is the equation of the tangent.
$$y=10x-6$$