How do you find the coordinates of the vertex #y= 2x^2 + 7x - 21 #?

Question

Hazel Anglin · Accepted Answer

Vertex is #(-7/4,-217/8)# or #(-1 3/4,-27 1/8)#

To find the coordinates of vertex of #y=2x^2+7x-21#, one should convert this equation into vertex form i.e.

#(y-k)=a(x-h)^2#, where vertex is #(h,k)#

Now #y=2x^2+7x-21#

#hArry=2(x^2+7/2x)-21#

#=2(x^2+2xx7/4xx x+(7/4)^2-(7/4)^2)-21#

#=2((x+7/4)^2-(7/4)^2)-21#

#=2(x+7/4)^2-2xx49/16-21#

#=2(x+7/4)^2-49/8-21#

#=2(x+7/4)^2-217/8#

or #(y+217/8)=2(x+7/4)^2#

Hence, vertex is #(-7/4,-217/8)# or #(-1 3/4,-27 1/8)#

graph{2x^2+7x-21 [-6, 4, -28.56, -8.56]}

Eli Assouline · Answer

undefined To find the coordinates of the vertex of the parabola represented by the equation $ y = 2x^2 + 7x - 21 $, you can use the formula for the x-coordinate of the vertex: $ x = \frac{{-b}}{{2a}} $, where $ a $ and $ b $ are the coefficients of the quadratic term and the linear term, respectively. Then, substitute the value of $ x $ into the equation to find the corresponding $ y $-coordinate. So, for the given equation, $ a = 2 $ and $ b = 7 $. Plugging these values into the formula, you get:

$$ x = \frac{{-7}}{{2 \cdot 2}} = -\frac{{7}}{{4}} $$

To find the $ y $-coordinate, substitute $ x = -\frac{{7}}{{4}} $ into the equation:

$$ y = 2 \left(-\frac{{7}}{{4}}\right)^2 + 7 \left(-\frac{{7}}{{4}}\right) - 21 $$

$$ y = 2 \cdot \frac{{49}}{{16}} - \frac{{49}}{{4}} - 21 $$

$$ y = \frac{{49}}{{8}} - \frac{{49}}{{4}} - 21 $$

$$ y = \frac{{49}}{{8}} - \frac{{98}}{{8}} - \frac{{168}}{{8}} $$

$$ y = -\frac{{217}}{{8}} $$

Therefore, the coordinates of the vertex are $ \left(-\frac{{7}}{{4}}, -\frac{{217}}{{8}}\right) $.